10.4: Communication Competence
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In this section, we will cover cognitive and behavioral skills to help us more effectively create the communication climate we want to experience, contextual nuances in interpersonal needs, and discuss the importance mindfulness and awareness.
Effective Communication: Cognitive and Behavioral Skills
Cognitive skills involve thinking about others and behavioral skills involve actionable things we can actually say and do. Specifically, we will cover three skills for more effective climate messages: empathy, quadruple thinking, and metacommunication.
Empathy
You may have heard empathy defined as the ability to (metaphorically) “put yourself in someone else’s shoes,” to feel what another may be feeling. This description is technically accurate on one level, but empathy is actually more complex. Our human capacity for empathy has three levels: cognitive, affective, and compassionate.
The first is cognitive and involves more thinking than feeling. A more appropriate metaphor for this level is “putting on someone else’s perception glasses,” to attempt to view a situation in the way someone else might view it. It requires thinking about someone else’s thinking, considering factors that make up someone’s unique perceptual schema, and trying to view a situation through that lens. For example, employees don’t always view things the way managers do. A good manager can see through employee glasses and anticipate how workplace actions, decisions, and/or messages may be interpreted.
The second level is affective, or emotional, and involves attempting to feel the emotions of others. The “shoes” metaphor fits best for this level. Attempting to truly feel what other humans feel requires envisioning exactly what they might be going through in their lives. Doing so effectively might even require “taking off your own shoes.” For example, to empathize with a complaining customer, we can temporarily put our own needs aside, and really picture what it would feel like to be the customer experiencing the problem situation. Your own need might be to take care of the complaint quickly so you can go to lunch. Yet, if it were you in the problem situation, you would likely want someone to be warm, attentive, and supportive, and take the time needed to solve the problem.
This level of empathy is often confused with sympathy, something with which you are probably already very familiar. The two are related but are not the same. Feeling sympathy means feeling bad for or sorry about something another person might be going through, but understanding and feeling it from your own perspective, through your own perception glasses, and in your own shoes. We all recognize that losing a pet is likely to be devastating for someone. We therefore feel sympathy for our friend because their dog died. However, feeling empathy requires making an effort to see the situation through their glasses and shoes. What this means is that we consider how they may see and feel the situation differently from us. For instance, we may have experienced many pet losses and even human losses in your life, so yet another pet loss may not feel that significant to us. But, if this is your friend’s first significant loss, they may likely feel more devastation than we would. We can respond more appropriately and with more warmth by letting go of our own perspective and attempting to see and feel the situation as they might. Another way to distinguish between sympathy and empathy is by seeing sympathy as “feeling for…” (as in feeling sorry for or feeling compassion for another person) and empathy as “feeling with…” as in actually feeling the emotions of another person.
The third level of empathy is the compassionate concern for the well-being of our fellow humans (Goleman, 2006). Feeling empathy at this level motivates us to act compassionately in the interest of others. Examples may include dropping off a casserole for a grieving friend, taking some of your coworker’s calls when they are especially busy or stressed, or organizing a neighborhood clean-up. With this level of empathy, we sense what people need and feel compelled to help. Most of us are usually able to empathize at this level with people who are important to us.
Strategies for building empathy: While empathy comes more naturally for some people than others, it is a skill that can be developed (Goleman, 2006) with a greater awareness of and attention to the perception process. Remember that perception is unique to each person. We all interpret and judge the world through our own set of perception glasses that are framed by factors such as upbringing, family background, ethnicity, age, attitude, knowledge of person and situation, past experiences, amount of exposure to others, social roles, etc.
Below addresses specific ways to build our empathy muscles. The strategies fall into two categories: adding information to the rims of our perception glasses and bringing attention to the perception process itself.
Add more information to our perception glasses. In order to add more information to our perception glasses, we need to find out what we can about a situation or person with whom we are seeking to understanding and empathize. We can do this by:
- Taking in information: When we observe, listen, question, perception check, paraphrase, pay attention to nonverbals and feelings, we take information in rather than putting information out (e.g., listening more and talking less).
- Broaden or narrow our perspective: Sometimes we feel stuck, allowing one interaction with one person to become all-consuming. If we remember how big the world is and how many people are dealing with similar situations right now, we gain perspective that helps us see the situation in a different way. On the other hand, sometimes we generalize too broadly, seeing an entire group of people in one way, or assuming all things are bad at our workplace. Focusing on one person or one situation a time is another way to helpfully shift perspectives.
- Imagine or seek stories and info (through books, films, articles, technology): We can learn and imagine what people’s lives are really like by reading, watching, or listening to the stories of others.
- Seek out actual experiences to help us understand what it’s like to be in others’ shoes: We can do something experiential like a ride-along with a police officer or spend a day on the streets to really try to feel what it’s like to be in a situation in which we are not familiar.
Bring attention to the perception process. Pull down our own perception glasses and try on a pair of someone else’s. Thinking about our thinking is a process called metacognition. By turning our attention toward the way we perceive information and how that perception makes us feel. What factors make up the rims of our glasses and how do these factors shape our perspectives, thoughts, feelings, and actions? Consider what makes another person unique, and what rim factors may influence the person’s perspectives and feelings. We should try to see the situation through those glasses, inferring how unique perceptual schemas might shape the others person’s emotions and actions too. Remember, though, we can never be certain how or why people do what they do. Only they know for sure. But communication can be more effective if we at least give some type of speculative forethought before we act or react. And when in doubt, we can always ask.
Climate-Centered Message Planning
Our second mindful competence skill is called Climate-Centered Message Planning (CCMP), which is a term coined by Gerber and Murphy (2019). CCMP refers to the conscious encoding (planning and forethought) involved in meeting communication goals. CCMP requires two steps and takes the basics of empathy a bit further into message construction. The steps include: 1. Think about what we want to say or do. What is our goal? What outcome(s) do we hope to achieve? What message or behaviors are we considering? What needs do we hope to fulfill? What emotional temperature do we hope to create? Which behaviors or message strategies will help us achieve it? 2. Think about how the other person (or persons) might hear (or perceive) what we say. Here, we should put on their perception glasses and consider as many factors as possible that affect how the person might see and feel our message. We should think about whether the message is likely to be perceived and received as intended. If not, rethink what we want to say so that they will be more likely hear what you want them to hear (so a person is more likely to interpret your messages as you intend it to be interpreted). Remember once again, we can never completely ensure that someone “hears what we want them to hear” (interprets what we intended). However, with some awareness and forethought, we can ensure there’s a better chance of it. CCMP also helps us with better awareness of how what we say and how we say it may impact another person’s relational or face needs. Our consideration of what human beings “need” will help us infer how they might react to messages emotionally, intellectually, or relationally. Doing so helps us communicate more effectively and appropriately whatever our goal may be.
Metacommunication
Our third skill is an action skill: the skill of metacommunication. It, too, requires mindfully elevating awareness beyond the content level of communication, but also requires us to actually discuss things such as needs and relational messages aloud. Metacommunication literally means communicating about communication, and occurs when we talk to each other about any part of the communication process, including what is said or done, how it is interpreted, how we feel, what we wish had been said or done, etc. For example, metacommunication occurs anytime you say “It’s frustrating when you interrupt me,” or “I wish you’d have asked me before you made that decision.” Other forms of metacommunication bring relational messages and social needs right to the surface level for discussion. For example, if you said “when you brought that up in front of my friends, I felt embarrassed and undignified,” or “when I don’t hear from you, I feel less connected to you.” Metacommunication can involve any of the skills we’ve learned so far (“I” messages, perception checking, etc.) and can be used deliberately to address our own wants, needs, or to clarify our intentions when something we’ve expressed may have been ill-received. Scholar and speaker Brene Brown recommends using phrases such as “the story I’m making up about this is…” to explain the way we perceived something and “help me better understand” as a form of listening to understand how another person may have perceived something. Metacommunication can help us in the middle of interactions to clarify and prevent misunderstandings as we both send and receive messages. For example, if you notice someone reacting in a way you didn’t intend, you can ask about it (“how are you feeling right now? What are you hearing me say?”) or you can clarify your intent and adjust (“My intent was not for you to feel disrespected. How can I say this differently so that you hear my respect for you?”). We can also respond to the cold relational messages of others with “When you say it that way, I hear not only what you’re saying but an extra message that you don’t think I’m capable” or “not giving me options leaves me feeling boxed in and I really want to feel more freedom in this relationship.”
Contextual Communication: Context and Needs
Context influences all of our communication, but it also has important implications when it comes to the interpersonal needs we have discussed throughout this chapter. Below we discuss how three types of contextual nuances that influence our needs: relational, individual, and co-cultural.
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Relational Context:
Based on the relational context, the relationship we have with others (friend, colleague, etc.), different people will meet different needs. We may expect a specific type of relationship to meet one need (but not the other) or place more or less importance on that relationship meeting a particular need. For example, in workplace relationships with colleagues, managers, or people we manage we may expect these people, and place higher value on these people meeting needs pertaining to competence, capability, control, etc. We may not expect these relationships to, or place lesser importance on them, meeting needs such as warmth and kindness. Conversely, we may expect intimate relationships with friends and romantic partners to satisfy needs related to connection, and not expect them to, or place less importance, on them meeting needs like control. -
Individual Context:
Individuals tend to vary in the level of desire for certain needs to be met. We may have low needs in one area and medium or high needs in another. For example, we may have a high need for connection and a low need for freedom. Conversely, another individual may have a low need for connection and high need for freedom. We also see a difference between how much needs are wanted (we want our needs met from others) and expressed (how much we want and do work toward meeting these needs in others). For example, we may have a need to feel respected but do little to show respect to another to meet their respect need. -
Co-Cultural Context:
Group membership and identity based on factors like gender, generation, social class, etc. can influence the value placed on a particular need. For example, men are often taught (through socialization) to strongly value the needs of autonomy, control, freedom from imposition by others, space, and privacy. Conversely, women are often taught to place higher value on needs of belonging, inclusion, warmth, and kindness. Because of this, it is important to recognize what we genuinely want in terms of needs- not just what we think we should want or were told we should value.
Reflective Communication: Mindfulness
The word mindfulness refers to “paying attention on purpose,” and has many uses in personal and work life. For interpersonal communication purposes, mindfulness relates to becoming more conscious of how we encode and decode messages. We can better meet our communication goals with increased awareness of how communication carries relational subtexts, how those subtexts may be perceived to meet (or not meet) social needs, and how those perceptions might result in a warm or cold emotional temperature. As with all communication competence skills, awareness helps us shift from a habitual or automatic state of being and thinking to a mindful and thoughtful state where we put more effort, attention and forethought into what we hope to accomplish and why.
Becoming mindful of climate means increasing awareness of the needs of self and others before, during, and after interactions. It requires reflecting on of our own desires, thought processes and emotional reactions, and with applied forethought, thinking about and speculating about those of others. Learning about relational messages and social needs gives us access to a greater variety of perceptual frameworks through which to view communication (e.g., how might this message be received by others?). It also requires that during interactions we observe, reflect on, and attend to others’ emotional reactions and shift gears midstream if necessary. For example, if mid-interaction we observe a person’s outward response that seems to indicate embarrassment, shame, agitation or defensiveness, we can adjust our behavior or discuss and clarify our intent. We may even take notice of an interaction after it occurred, reviewing it and considering how well it went or how we might do better next time. Through awareness, reflection, mindfulness we can build a cognitively complex repertoire of skill, knowledge, and motivation that helps us engage in a skillful dance of communication that attempts to honor social needs.
Practice
Consider what you might say and do to convey the given relational subtext.
- How might you decline a friend’s invitation to coffee with respect and affection?
- How could you address an employee’s need for autonomy when telling her she is needed to work this weekend? The need for dignity?
- How might you disagree with a classmate’s politics by recognizing their need for inclusion? Respect? The need to matter?
- How might you address the autonomy or affection needs of your significant other when you request that they come home on time tonight?


10.1: An Introduction to Probability
We continue with an informal discussion intended to motivate the more structured development that will follow. Consider the “spinner” shown in Figure 10.1 . Suppose we give it a good thwack so that the arrow goes round and round. We then record the number of the region in which the pointer comes to rest. Then observers, none of whom have studied combinatorics, might make the following comments:
- The odds of landing in region 1 are the same as those for landing in region 3.
- You are twice as likely to land in region 2 as in region 4.
- When you land in an odd numbered region, then 60% of the time, it will be in region 5.
We will now develop a more formal framework that will enable us to make such discussions far more precise. We will also see whether Alice is being entirely fair to Bob in her proposed game to one hundred.
We begin by defining a probability space as a pair \((S,P)\) where \(S\) is a finite set and \(P\) is a function that whose domain is the family of all subsets of \(S\) and whose range is the set [0,1] of all real numbers which are non-negative and at most one. Furthermore, the following two key properties must be satisfied:
- \(P(\emptyset) = 0\) and \(P(S) = 1\).
- If \(A\) and \(B\) are subsets of \(S\), and \(A \cap B= \emptyset\), then \(P(A \cup B) = P(A) + P(B)\).
When \((S,P)\) is a probability space, the function \(P\) is called a probability measure , the subsets of \(S\) are called events , and when \(E⊆S\), the quantity \(P(E)\) is referred to as the probability of the event \(E\).
Note that we can consider \(P\) to be extended to a mapping from \(S\) to [0,1] by setting \(P(x)=P(\{x\})\) for each element \(x \in S\). We call the elements of \(S\) outcomes (some people prefer to say the elements are elementary outcomes ) and the quantity \(P(x)\) is called the probability of \(x\). It is important to realize that if you know \(P(x)\) for each \(x \in S\), then you can calculate \(P(E)\) for any event \(E\), since (by the second property), \(P(E) = \sum_{x \in X} P(x)\).
For the spinner, we can take \(S=\{1,2,3,4,5\}\), with \(P(1)=P(3)=P(4)=1/8, P(2)=2/8=1/4\) and \(P(5)=3/8\). So \(P(\{2,3\}) = 1/8 + 2/8 = 3/8\).
Let \(S\) be a finite, nonempty set and let \(n=|S|\). For each \(E⊆S\), set \(P(E)=|E|/n\). In particular, \(P(x)=1/n\) for each element \(x \in S\). In this trivial example, all outcomes are equally likely.
If a single six sided die is rolled and the number of dots on the top face is recorded, then the ground set is \(S=\{1,2,3,4,5,6\}\) and \(P(i)=1/6\) for each \(i \in S\). On the other hand, if a pair of dice are rolled and the sum of the dots on the two top faces is recorded, then \(S=\{2,3,4,…,11,12\}\) with \(P(2)=P(12)=1/36, P(3)=P(11)=2/36, P(4)=P(10)=3/36, P(5)=P(9)=4/36, P(6)=P(8)=5/36\) and \(P(7)=6/36\). To see this, consider the two die as distinguished, one die red and the other one blue.  So each has probability 1/36. Then, for example, there are three pairs that yield a total of four, namely (3,1), (2,2) and (1,3). So the probability of rolling a four is 3/36=1/12.
In Alice's game as described above, the set \(S\) can be {0,1,2,3,4,5}, the set of possible differences when a pair of dice are rolled. In this game, we will see that the correct definition of the function \(P\) will set \(P(0)=6/36; P(1)=10/36; P(2)=8/36; P(3)=6/36; P(4)=4/36\); and \(P(5)=2/36\). Using Xing's more compact notation, we could say that \(P(0)=1/6\) and \(P(d)=2(6−d)/36\) when \(d > 0\).
A jar contains twenty marbles, of which six are red, nine are blue and the remaining five are green. Three of the twenty marbles are selected at random. Let \(X=\{0,1,2,3,4,5\}\), and for each \(x \in X\), let \(P(x)\) denote the probability that the number of blue marbles among the three marbles selected is \(x\). Then \(P(i)=C(9,i)C(11,3−i)/C(20,3)\) for \(i=0,1,2,3\), while \(P(4)=P(5)=0\). Bob says that it doesn't make sense to have outcomes with probability zero, but Carlos says that it does.
In some cards games, each player receives five cards from a standard deck of 52 cards—four suits (spades, hearts, diamonds and clubs) with 13 cards, ace though king in each suit. A player has a full house if there are two values \(x\) and \(y\) for which he has three of the four \(x\)'s and two of the four \(y\)'s, e.g. three kings and two eights. If five cards are drawn at random from a standard deck, the probability of a full house is
\(\dfrac{\binom{13}{1} \binom{12}{1} \binom{4}{3} \binom{4}{2}}{\binom{52}{5}} \approx 0.00144\)


8.4.4: Safe Microwave Defrosting and Cooking
Many people use a microwave to cook or reheat food. It is important for the Home Health Aide/Personal Care Aide to follow safety precautions when using a microwave for food preparation. The United States Department of Agriculture has specific guidelines for safe use of a microwave for food preparation. You can read more about the following guidelines based on the USDA (2013b) recommendations for safe microwave use at: http :// www . fsis . usda . gov / wps / portal / fsis / topics / food – safety – education / get – answers / food – safety – fact – sheets / appliances – and – thermometers / cooking – safely – in – the – microwave / cooking – safely – in – the – microwave – oven
- When heating or cooking foods in a microwave, always use microwave safe dishes. Never use plastic or foam trays. These items can warp, causing harmful chemicals to spill into the food.
- Cover food with microwave safe covers. Loosen the lid slightly to allow steam to escape to prevent burns. This allows moisture to stay inside which provides for even heating. Unevenly heated or cooked foods can have areas where bacteria live.
- When defrosting, remove packaging food came in. Do not use plastic or foam trays.
- Cook food immediately after defrosting. Do not defrost and save the food to cook later.
- Cook large cuts of meat at 50% power for a longer time rather than at 100% power. This allows enough time for heat to reach the center without overcooking outer areas.
- Stir and rotate food halfway during cooking or reheating to avoid cold spots where bacteria can live.
- Always allow foods cooked in a microwave to stand for a couple of minutes to avoid being burned by steam.
- Always ensure reheated foods have reached a temperature of 165 degrees F. Use a thermometer to check for doneness.
Safe Cooking Temperatures
The USDA has guidelines about safe food cooking temperatures. To avoid food-borne illnesses, these should be followed. Home Health Aides/Personal Care Aides should always check food for doneness with a thermometer . Just because food appears to be cooked from the way it looks on the outside does not mean it is cooked in the center. Insert a thermometer into the thickest part of the meat or the center of the food. Do not allow the thermometer to touch bone, fat, or gristle (United States Department of Agriculture, 2011).
| Type of Food | Temperature to which to reheat |
|---|---|
| Beef , pork , lamb , and veal steaks , chops , or roasts | 145 degrees F |
| Ground beef , pork , lamb , veal | 160 degrees F |
| Poultry | 165 degrees F |
| Egg dishes | 160 degrees F |
| Reheating leftovers | 165 degrees F |
Source: United States Department of Agriculture, 2011
True or False
1. If a food is brown on the outside, it is done. True or False? ___________
Multiple Choice
2. At which temperature should poultry be cooked to?
a). 165 degrees
b). 145 degrees
c). 160 degrees
d). 155 degrees
3. At which temperature should ground beef be cooked to?
a). 165 degrees
b). 145 degrees
c). 160 degrees
d). 155 degrees
4. At which temperature should leftovers be reheated to?
a). 165 degrees
b). 145 degrees
c). 160 degrees
d). 155 degrees
5. At which temperature should steaks, chops, and roasts be cooked to?
a). 165 degrees
b). 145 degrees
c). 160 degrees
d). 155 degrees
- Answer
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1. False
2. A
3. C
4. A
5. B
FEEDBACK :
1. Just because food appears to be cooked from the way it looks on the outside does not mean it is cooked in the center. Use a thermometer to test for doneness.
2. Poultry should be cooked to 165 degrees.
3. Ground beef should be cooked to 160 degrees.
4. Leftovers should be reheated to 165 degrees.
5. Steaks, chops, and roasts should be cooked to 145 degrees.


7.2: Solve Quadratic Equations Using the Square Root Property
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Learning Objectives
By the end of this section, you will be able to:
- Solve quadratic equations of the form \(ax^{2}=k\) using the Square Root Property
- Solve quadratic equations of the form \(a(x–h)^{2}=k\) using the Square Root Property
Before you get started, take this readiness quiz.
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Simplify: \(\sqrt{128}\).
If you missed this problem, review Example 8.13. -
Simplify: \(\sqrt{\frac{32}{5}}\).
If you missed this problem, review Example 8.50. -
Factor: \(9 x^{2}-12 x+4\).
If you missed this problem, review Example 6.23.
A quadratic equation is an equation of the form \(a x^{2}+b x+c=0\) , where \(a≠0\). Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form \(ax^{2}\). We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.
We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.
Solve Quadratic Equations of the Form \(ax^{2}=k\) using the Square Root Property
We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation \(x^{2}=9\).
\(x^{2}=9\)
Put the equation in standard form.
\(x^{2}-9=0\)
Factor the difference of squares.
\((x-3)(x+3)=0\)
Use the Zero Produce Property.
\(x-3=0 \quad x-3=0\)
Solve each equation.
\(x=3 \quad x=-3\)
We can easily use factoring to find the solutions of similar equations, like \(x^{2}=16\) and \(x^{2}=25\), because \(16\) and \(25\) are perfect squares. In each case, we would get two solutions, \(x=4, x=-4\) and \(x=5, x=-5\)
But what happens when we have an equation like \(x^{2}=7\)? Since \(7\) is not a perfect square, we cannot solve the equation by factoring.
Previously we learned that since \(169\) is the square of \(13\), we can also say that \(13\) is a square root of \(169\). Also, \((-13)^{2}=169\), so \(−13\) is also a square root of \(169\). Therefore, both \(13\) and \(−13\) are square roots of \(169\). So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:
If \(n^{2}=m\), then \(n\) is a square root of \(m\).
Since these equations are all of the form \(x^{2}=k\), the square root definition tells us the solutions are the two square roots of \(k\). This leads to the Square Root Property .
Definition \(\PageIndex{1}\)
Square Root Property
If \(x^{2}=k\), then
\(x=\sqrt{k} \quad\) or \(\quad x=-\sqrt{k} \quad\) or \(\quad x=\pm \sqrt{k}\)
Notice that the Square Root Property gives two solutions to an equation of the form \(x^{2}=k\), the principal square root of \(k\) and its opposite. We could also write the solution as \(x=\pm \sqrt{k}\). We read this as \(x\) equals positive or negative the square root of \(k\).
Now we will solve the equation \(x^{2}=9\) again, this time using the Square Root Property.
\(\begin{aligned} &x^{2} =9 \\ \text { Use the Square Root Property. } \quad& x=\pm \sqrt{9} \\& x =\pm 3 \end{aligned}\)
So \(x=3\) or \(x=-3\)
What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation \(x^{2}=7\).
\(x^{2}=7\)
Use the Square Root Property. \(x=\sqrt{7}, \quad x=-\sqrt{7}\)
We cannot simplify \(\sqrt{7}\), so we leave the answer as a radical.
Example \(\PageIndex{1}\) How to Solve a Quadratic Equation of the form \(ax^{2}-k\) Using the Square Root Property
Solve: \(x^{2}-50=0\).
Solution :
| Step 1 : Isolate the quadratic term and make its coefficient one. | Add \(50\) to both sides to get \(x^{2}\) by itself. | \(\begin{aligned} x^{2}-50 &=0 \\ x^{2} &=50 \end{aligned}\) |
| Step 2 : Use the Square Root Property. | Remember to write the \(\pm\) symbol. | \(x=\pm \sqrt{50}\) |
| Step 3 : Simplify the radical. | Rewrite to show two solutions. | \(\begin{array}{l}{x=\pm \sqrt{25} \cdot \sqrt{2}} \\ {x=\pm 5 \sqrt{2}} \\ {}x=5\sqrt{2}, \:x=-5\sqrt{2}\end{array}\) |
| Step 4 : Check the solutions. | Substitute in \(x=5 \sqrt{2}\) and \(x=-5 \sqrt{2}\) |
\(\begin{array}{r}{x^{2}-50=0} \\ {(\color{red}{5 \sqrt{2}}\color{black}{)}^{2}-50 \stackrel{?}{=} 0} \\ {25 \cdot 2-50 \stackrel{?}{=} 0} \\ {0=0}\end{array}\) \(\begin{array}{r}{x^{2}-50=0} \\ {(\color{red}{-5 \sqrt{2}}\color{black}{)}^{2}-50 \stackrel{?}{=} 0} \\ {25 \cdot 2-50 \stackrel{?}{=} 0} \\ {0=0}\end{array}\) |
Exercise \(\PageIndex{1}\)
Solve: \(x^{2}-48=0\).
- Answer
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\(x=4 \sqrt{3}, x=-4 \sqrt{3}\)
Exercise \(\PageIndex{2}\)
Solve: \(y^{2}-27=0\).
- Answer
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\(y=3 \sqrt{3}, y=-3 \sqrt{3}\)
The steps to take to use the Square Root Property to solve a quadratic equation are listed here.
Solve a Quadratic Equation Using the Square Root Property
- Isolate the quadratic term and make its coefficient one.
- Use Square Root Property.
- Simplify the radical.
- Check the solutions.
In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient \(3\) before using the Square Root Property.
Example \(\PageIndex{2}\)
Solve: \(3 z^{2}=108\).
Solution :
| \(3 z^{2}=108\) | |
| The quadratic term is isolated. Divide by \(3\) to make its coefficient \(1\). | \(\frac{3 z^{2}}{3}=\frac{108}{3}\) |
| Simplify. | \(z^{2}=36\) |
| Use the Square Root Property. | \(z=\pm \sqrt{36}\) |
| Simplify the radical. | \(z=\pm 6\) |
| Rewrite to show two solutions. | \(z=6, \quad z=-6\) |
|
Check the solutions: |
Exercise \(\PageIndex{3}\)
Solve: \(2x^{2}=98\).
- Answer
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\(x=7, x=-7\)
Exercise \(\PageIndex{4}\)
Solve: \(5m^{2}=80\).
- Answer
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\(m=4, m=-4\)
The Square Root Property states ‘If \(x^{2}=k\),’ What will happen if \(k<0\)? This will be the case in the next example.
Example \(\PageIndex{3}\)
Solve: \(x^{2}+72=0\).
Solution :
| \(x^{2}+72=0\) | |
| Isolate the quadratic term. | \(x^{2}=-72\) |
| Use the Square Root Property. | \(x=\pm \sqrt{-72}\) |
| Simplify using complex numbers. | \(x=\pm \sqrt{72} i\) |
| Simplify the radical. | \(x=\pm 6 \sqrt{2} i\) |
| Rewrite to show two solutions | \(x=6 \sqrt{2} i, x=-6 \sqrt{2} i\) |
|
Check the solutions:
\(\begin{array}{r}{x^{2}+72=0}
|
\(\begin{array}{r}{x^{2}+72=0}
|
Exercise \(\PageIndex{5}\)
Solve: \(c^{2}+12=0\).
- Answer
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\(c=2 \sqrt{3} i, \quad c=-2 \sqrt{3} i\)
Exercise \(\PageIndex{6}\)
Solve: \(q^{2}+24=0\).
- Answer
-
\(c=2 \sqrt{6} i, \quad c=-2 \sqrt{6} i\)
Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.
Example \(\PageIndex{4}\)
Solve: \(\frac{2}{3} u^{2}+5=17\).
Solution :
| \(\frac{2}{3} u^{2}+5=17\) | |
| Isolate the quadratic term. | |
| Multiply by \(\frac{3}{2}\) to make the coefficient \(1\). | |
| Simplify. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Simplify. | |
| Rewrite to show two solutions. | |
|
Check:
\(\begin{array}{r}{\frac{2}{3}u^{2}+5=17}
|
\(\begin{array}{r}{\frac{2}{3}u^{2}+5=17}
|
Exercise \(\PageIndex{7}\)
Solve: \(\frac{1}{2} x^{2}+4=24\).
- Answer
-
\(x=2 \sqrt{10}, x=-2 \sqrt{10}\)
Exercise \(\PageIndex{8}\)
Solve: \(\frac{3}{4} y^{2}-3=18\).
- Answer
-
\(y=2 \sqrt{7}, y=-2 \sqrt{7}\)
The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .
Example \(\PageIndex{5}\)
Solve: \(2 x^{2}-8=41\).
Solution :
| Isolate the quadratic term. | |
| Divide by \(2\) to make the coefficient \(1\). | |
| Simplify. | |
| Use the Square Root Property. | |
| Rewrite the radical as a fraction of square roots. | |
| Rationalize the denominator. | |
| Simplify. | |
| Rewrite to show two solutions. | |
|
Check: We leave te check for you. |
Exercise \(\PageIndex{9}\)
Solve: \(5 r^{2}-2=34\).
- Answer
-
\(r=\frac{6 \sqrt{5}}{5}, \quad r=-\frac{6 \sqrt{5}}{5}\)
Exercise \(\PageIndex{10}\)
Solve: \(3 t^{2}+6=70\).
- Answer
-
\(t=\frac{8 \sqrt{3}}{3}, \quad t=-\frac{8 \sqrt{3}}{3}\)
Solve Quadratic Equation of the Form \(a(x-h)^{2}=k\) Using the Square Root Property
We can use the Square Root Property to solve an equation of the form \(a(x-h)^{2}=k\) as well. Notice that the quadratic term, \(x\), in the original form \(ax^{2}=k\) is replaced with \((x-h)\).
The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of \(a\), then the Square Root Property can be used on \((x-h)^{2}\).
Example \(\PageIndex{6}\)
Solve: \(4(y-7)^{2}=48\).
Solution :
| \(4(y-7)^{2}=48\) | |
| Divide both sides by the coefficient \(4\). | \((y-7)^{2}=12\) |
| Use the Square Root Property on the binomial. | \(y-7=\pm \sqrt{12}\) |
| Simplify the radical. | \(y-7=\pm 2 \sqrt{3}\) |
| Solve for \(y\). | \(y=7 \pm 2 \sqrt{3}\) |
| Rewrite to show two solutions. |
\(y=7+2 \sqrt{3}\)
\(y=7-2 \sqrt{3}\) |
|
Check: |
Exercise \(\PageIndex{11}\)
Solve: \(3(a-3)^{2}=54\).
- Answer
-
\(a=3+3 \sqrt{2}, \quad a=3-3 \sqrt{2}\)
Exercise \(\PageIndex{12}\)
Solve: \(2(b+2)^{2}=80\).
- Answer
-
\(b=-2+2 \sqrt{10}, \quad b=-2-2 \sqrt{10}\)
Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.
Example \(\PageIndex{7}\)
Solve: \(\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}\).
Solution :
\(\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}\)
Use the Square Root Property.
\(x-\frac{1}{3}=\pm \sqrt{\frac{5}{9}}\)
Rewrite the radical as a fraction of square roots.
\(x-\frac{1}{3}=\pm \frac{\sqrt{5}}{\sqrt{9}}\)
Simplify the radical.
\(x-\frac{1}{3}=\pm \frac{\sqrt{5}}{3}\)
Solve for \(x\).
\(x=\frac{1}{3} \pm \frac{\sqrt{5}}{3}\)
Rewrite to show two solutions.
\(x=\frac{1}{3}+\frac{\sqrt{5}}{3}, x=\frac{1}{3}-\frac{\sqrt{5}}{3}\)
Check:
We leave the check for you.
Exercise \(\PageIndex{13}\)
Solve: \(\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}\).
- Answer
-
\(x=\frac{1}{2}+\frac{\sqrt{5}}{2}, x=\frac{1}{2}-\frac{\sqrt{5}}{2}\)
Exercise \(\PageIndex{14}\)
Solve: \(\left(y+\frac{3}{4}\right)^{2}=\frac{7}{16}\).
- Answer
-
\(y=-\frac{3}{4}+\frac{\sqrt{7}}{4}, y=-\frac{3}{4}-\frac{\sqrt{7}}{4}\)
We will start the solution to the next example by isolating the binomial term.
Example \(\PageIndex{8}\)
Solve: \(2(x-2)^{2}+3=57\).
Solution :
\(2(x-2)^{2}+3=57\)
Subtract \(3\) from both sides to isolate the binomial term.
\(2(x-2)^{2}=54\)
Divide both sides by \(2\).
\((x-2)^{2}=27\)
Use the Square Root Property.
\(x-2=\pm \sqrt{27}\)
Simplify the radical.
\(x-2=\pm 3 \sqrt{3}\)
Solve for \(x\).
\(x=2 \pm 3 \sqrt{3}\)
Rewrite to show two solutions.
\(x=2+3 \sqrt{3}, x=2-3 \sqrt{3}\)
Check:
We leave the check for you.
Exercise \(\PageIndex{15}\)
Solve: \(5(a-5)^{2}+4=104\).
- Answer
-
\(a=5+2 \sqrt{5}, a=5-2 \sqrt{5}\)
Exercise \(\PageIndex{16}\)
Solve: \(3(b+3)^{2}-8=88\).
- Answer
-
\(b=-3+4 \sqrt{2}, \quad b=-3-4 \sqrt{2}\)
Sometimes the solutions are complex numbers.
Example \(\PageIndex{9}\)
Solve: \((2 x-3)^{2}=-12\).
Solution :
\((2 x-3)^{2}=-12\)
Use the Square Root Property.
\(2 x-3=\pm \sqrt{-12}\)
Simplify the radical.
\(2 x-3=\pm 2 \sqrt{3} i\)
Add \(3\) to both sides.
\(2 x=3 \pm 2 \sqrt{3} i\)
Divide both sides by \(2\).
\(x=\frac{3 \pm 2 \sqrt{3 i}}{2}\)
Rewrite in standard form.
\(x=\frac{3}{2} \pm \frac{2 \sqrt{3} i}{2}\)
Simplify.
\(x=\frac{3}{2} \pm \sqrt{3} i\)
Rewrite to show two solutions.
\(x=\frac{3}{2}+\sqrt{3} i, x=\frac{3}{2}-\sqrt{3} i\)
Check:
We leave the check for you.
Exercise \(\PageIndex{17}\)
Solve: \((3 r+4)^{2}=-8\).
- Answer
-
\(r=-\frac{4}{3}+\frac{2 \sqrt{2} i}{3}, r=-\frac{4}{3}-\frac{2 \sqrt{2} i}{3}\)
Exercise \(\PageIndex{18}\)
Solve: \((2 t-8)^{2}=-10\).
- Answer
-
\(t=4+\frac{\sqrt{10} i}{2}, t=4-\frac{\sqrt{10 i}}{2}\)
The left sides of the equations in the next two examples do not seem to be of the form \(a(x-h)^{2}\). But they are perfect square trinomials, so we will factor to put them in the form we need.
Example \(\PageIndex{10}\)
Solve: \(4 n^{2}+4 n+1=16\).
Solution :
We notice the left side of the equation is a perfect square trinomial. We will factor it first.
| \(4 n^{2}+4 n+1=16\) | |
| Factor the perfect square trinomial. | \((2 n+1)^{2}=16\) |
| Use the Square Root Property. | \(2 n+1=\pm \sqrt{16}\) |
| Simplify the radical. | \(2 n+1=\pm 4\) |
| Solve for \(n\). | \(2 n=-1 \pm 4\) |
| Divide each side by \(2\). | \(\begin{aligned} \frac{2 n}{2} &=\frac{-1 \pm 4}{2} \\ n &=\frac{-1 \pm 4}{2} \end{aligned}\) |
| Rewrite to show two solutions. | \(n=\frac{-1+4}{2}, n=\frac{-1-4}{2}\) |
| Simplify each equation. | \(n=\frac{3}{2}, \quad n=-\frac{5}{2}\) |
|
Check: |
Exercise \(\PageIndex{19}\)
Solve: \(9 m^{2}-12 m+4=25\).
- Answer
-
\(m=\frac{7}{3}, \quad m=-1\)
Exercise \(\PageIndex{20}\)
Solve: \(16 n^{2}+40 n+25=4\).
- Answer
-
\(n=-\frac{3}{4}, \quad n=-\frac{7}{4}\)
Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.
Key Concepts
-
Square Root Property
- If \(x^{2}=k\), then \(x=\sqrt{k}\) or \(x=-\sqrt{k}\)or \(x=\pm \sqrt{k}\)
- Isolate the quadratic term and make its coefficient one.
- Use Square Root Property.
- Simplify the radical.
- Check the solutions.


9: Molecular Polarity, Intermolecular Forces and Solubility
-
- 9.2: Intermolecular Forces
- As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase “intermolecular attraction” to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.
Thumbnail: Diagram illustrating the polarity of a general carbon-fluorine bond, with a partial positive charge on carbon (δ+) and a partial negative charge of fluorine (δ−). Representation of Cδ+−Fδ− . (public domain, 4.0; Ben Mills via Wikimedia )


12.1: The Ellipse
- Write equations of ellipses in standard form.
- Graph ellipses centered at the origin.
- Graph ellipses not centered at the origin.
- Solve applied problems involving ellipses.
Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in Figure \(\PageIndex{1}\), is such a room. It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper.
Figure \(\PageIndex{1}\) : The National Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr)
Writing Equations of Ellipses in Standard Form
A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\)
Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points \((x,y)\) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus(plural: foci).
We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See Figure \(\PageIndex{3}\).
Figure \(\PageIndex{3}\)
Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci (Figure \(\PageIndex{4}\)).
Figure \(\PageIndex{4}\)
In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the \(x\)- and \(y\)-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.
To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.
Deriving the Equation of an Ellipse Centered at the Origin
To derive the equation of an ellipse centered at the origin, we begin with the foci \((−c,0)\) and \((c,0)\). The ellipse is the set of all points \((x,y)\) such that the sum of the distances from \((x,y)\) to the foci is constant, as shown in Figure \(\PageIndex{5}\).
Figure \(\PageIndex{5}\)
If \((a,0)\) is a vertex of the ellipse, the distance from \((−c,0)\) to \((a,0)\) is \(a−(−c)=a+c\). The distance from \((c,0)\) to \((a,0)\) is \(a−c\). The sum of the distances from the foci to the vertex is
\((a+c)+(a−c)=2a\)
If \((x,y)\) is a point on the ellipse, then we can define the following variables:
- \(d_1=\) the distance from \((−c,0)\) to \((x,y)\)
- \(d_2=\) the distance from \((c,0)\) to \((x,y)\)
By the definition of an ellipse, \(d_1+d_2\) is constant for any point \((x,y)\) on the ellipse. We know that the sum of these distances is \(2a\) for the vertex \((a,0)\). It follows that \(d_1+d_2=2a\) for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic.
\[\begin{align*} d_1+d_2&= 2a\\ \sqrt{{(x-(-c))}^2+{(y-0)}^2}+\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\qquad \text{Distance formula}\\ \sqrt{{(x+c)}^2+y^2}+\sqrt{{(x-c)}^2+y^2}&=2a\qquad \text{Simplify expressions.}\\ \sqrt{{(x+c)}^2+y^2}&=2a-\sqrt{{(x-c)}^2+y^2}\qquad \text{Move radical to opposite side.}\\ {(x+c)}^2+y^2&={\left[2a-\sqrt{{(x-c)}^2+y^2}\right]}^2\qquad \text{Square both sides.}\\ x^2+2cx+c^2+y^2&=4a^2-4a\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\qquad \text{Expand the squares.}\\ x^2+2cx+c^2+y^2&=4a^2-4a\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\qquad \text{Expand remaining squares.}\\ 2cx&=4a^2-4a\sqrt{{(x-c)}^2+y^2}-2cx\qquad \text{Combine like terms.}\\ 4cx-4a^2&=-4a\sqrt{{(x-c)}^2+y^2}\qquad \text{Isolate the radical.}\\ cx-a^2&=-a\sqrt{{(x-c)}^2+y^2}\qquad \text{Divide by 4.}\\ {\left[ cx-a^2\right]}^2&=a^2{\left[ \sqrt{{(x-c)}^2+y^2}\right] }^2\qquad \text{Square both sides.}\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\qquad \text{Expand the squares.}\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\qquad \text{Distribute } a^2\\ a^2x^2-c^2x^2+a^2y^2&=a^4-a^2c^2\qquad \text{Rewrite.}\\ x^2(a^2-c^2)+a^2y^2&=a^2(a^2-c^2)\qquad \text{Factor common terms.}\\ x^2b^2+a^2y^2&=a^2b^2\qquad \text{Set } b^2=a^2-c^2\\ \dfrac{x^2b^2}{a^2b^2}+\dfrac{a^2y^2}{a^2b^2}&=\dfrac{a^2b^2}{a^2b^2}\qquad \text{Divide both sides by } a^2b^2\\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}&=1\qquad \text{Simplify} \end{align*}\]
Thus, the standard equation of an ellipse is \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).This equation defines an ellipse centered at the origin. If \(a>b\),the ellipse is stretched further in the horizontal direction, and if \(b>a\), the ellipse is stretched further in the vertical direction.
Writing Equations of Ellipses Centered at the Origin in Standard Form
Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena.
The key features of the ellipse are its center, vertices , co-vertices , foci , and lengths and positions of the major and minor axes . Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse.
The standard form of the equation of an ellipse with center \((0,0)\) and major axis on the \(x\) -axis is
\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\]
where
- \(a>b\)
- the length of the major axis is \(2a\)
- the coordinates of the vertices are \((\pm a,0)\)
- the length of the minor axis is \(2b\)
- the coordinates of the co-vertices are \((0,\pm b)\)
- the coordinates of the foci are \((\pm c,0)\), where \(c^2=a^2−b^2\). See Figure \(\PageIndex{6a}\).
The standard form of the equation of an ellipse with center \((0,0)\) and major axis on the \(y\) -axis is
\[\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\]
where
- \(a>b\)
- the length of the major axis is \(2a\)
- the coordinates of the vertices are \((0,\pm a)\)
- the length of the minor axis is \(2b\)
- the coordinates of the co-vertices are \((\pm b,0)\)
- the coordinates of the foci are \((0,\pm c)\), where \(c^2=a^2−b^2\). See Figure \(\PageIndex{6b}\).
Note that the vertices, co-vertices, and foci are related by the equation \(c^2=a^2−b^2\). When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form.
Figure \(\PageIndex{6}\): (a) Horizontal ellipse with center \((0,0)\) (b) Vertical ellipse with center \((0,0)\)
-
Determine whether the major axis lies on the
x
- or
y
-axis.
- If the given coordinates of the vertices and foci have the form \((\pm a,0)\) and \((\pm c,0)\) respectively, then the major axis is the x -axis. Use the standard form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)
- If the given coordinates of the vertices and foci have the form \((0,\pm a)\) and \((\pm c,0)\),respectively, then the major axis is the y -axis. Use the standard form \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\)
- Use the equation \(c^2=a^2−b^2\), along with the given coordinates of the vertices and foci, to solve for \(b^2\).
- Substitute the values for \(a^2\) and \(b^2\) into the standard form of the equation determined in Step 1.
What is the standard form equation of the ellipse that has vertices \((\pm 8,0)\) and foci \((\pm 5,0)\)?
Solution
The foci are on the \(x\)-axis, so the major axis is the \(x\)-axis. Thus, the equation will have the form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)
The vertices are \((\pm 8,0)\),so \(a=8\) and \(a^2=64\).
The foci are \((\pm 5,0)\),so \(c=5\) and \(c^2=25\).
We know that the vertices and foci are related by the equation \(c^2=a^2−b^2\). Solving for \(b^2\), we have:
\[\begin{align*} c^2&=a^2-b^2\\ 25&=64-b^2\qquad \text{Substitute for } c^2 \text{ and } a^2\\ b^2&=39\qquad \text{Solve for } b^2 \end{align*}\]
Now we need only substitute \(a^2=64\) and \(b^2=39\) into the standard form of the equation. The equation of the ellipse is \(\dfrac{x^2}{64}+\dfrac{y^2}{39}=1\).
What is the standard form equation of the ellipse that has vertices \((0,\pm 4)\) and foci \((0,\pm \sqrt{15})\)?
- Answer
-
\(x^2+\dfrac{y^2}{16}=1\)
Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex?
Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form \((\pm a,0)\) or \((0, \pm a)\). Similarly, the coordinates of the foci will always have the form \((\pm c,0)\) or \((0, \pm c)\). Knowing this, we can use \(a\) and \(c\) from the given points, along with the equation \(c^2=a^2−b^2\), to find \(b^2\).
Writing Equations of Ellipses Not Centered at the Origin
Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated \(h\) units horizontally and \(k\) units vertically, the center of the ellipse will be \((h,k)\). This translation results in the standard form of the equation we saw previously, with \(x\) replaced by \((x−h)\) and y replaced by \((y−k)\).
The standard form of the equation of an ellipse with center \((h, k)\) and major axis parallel to the \(x\)-axis is
\[\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\]
where
- \(a>b\)
- the length of the major axis is \(2a\)
- the coordinates of the vertices are \((h\pm a,k)\)
- the length of the minor axis is \(2b\)
- the coordinates of the co-vertices are \((h,k\pm b)\)
- the coordinates of the foci are \((h\pm c,k)\),where \(c^2=a^2−b^2\). See Figure \(\PageIndex{7a}\).
The standard form of the equation of an ellipse with center \((h,k)\) and major axis parallel to the \(y\)-axis is
\[\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1\]
where
- \(a>b\)
- the length of the major axis is \(2a\)
- the coordinates of the vertices are \((h,k\pm a)\)
- the length of the minor axis is \(2b\)
- the coordinates of the co-vertices are \((h\pm b,k)\)
- the coordinates of the foci are \((h,k\pm c)\), where \(c^2=a^2−b^2\). See Figure \(\PageIndex{7b}\).
Just as with ellipses centered at the origin, ellipses that are centered at a point \((h,k)\) have vertices, co-vertices, and foci that are related by the equation \(c^2=a^2−b^2\). We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.
Figure \(\PageIndex{7}\): (a) Horizontal ellipse with center \((h,k)\) (b) Vertical ellipse with center \((h,k)\)
-
Determine whether the major axis is parallel to the \(x\)- or \(y\)-axis.
- If the y -coordinates of the given vertices and foci are the same, then the major axis is parallel to the \(x\)-axis. Use the standard form \(\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\)
- If the x -coordinates of the given vertices and foci are the same, then the major axis is parallel to the y -axis. Use the standard form \(\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1\)
- Identify the center of the ellipse \((h,k)\) using the midpoint formula and the given coordinates for the vertices.
- Find \(a^2\) by solving for the length of the major axis, \(2a\), which is the distance between the given vertices.
- Find \(c^2\) using \(h\) and \(k\), found in Step 2, along with the given coordinates for the foci.
- Solve for \(b^2\) using the equation \(c^2=a^2−b^2\).
- Substitute the values for \(h\), \(k\), \(a^2\), and \(b^2\) into the standard form of the equation determined in Step 1.
What is the standard form equation of the ellipse that has vertices \((−2,−8)\) and \((−2,2)\) and foci \((−2,−7)\) and \((−2,1)\)?
Solution
The \(x\)-coordinates of the vertices and foci are the same, so the major axis is parallel to the \(y\)-axis. Thus, the equation of the ellipse will have the form
\(\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1 \nonumber\)
First, we identify the center, \((h,k)\). The center is halfway between the vertices, \((−2,−8)\) and \((−2,2)\). Applying the midpoint formula, we have:
\[\begin{align} (h,k) &=\left(\dfrac{−2+(−2)}{2},\dfrac{−8+2}{2}\right) \nonumber \\ &=(−2,−3) \nonumber \end{align} \nonumber\]
Next, we find \(a^2\). The length of the major axis, \(2a\), is bounded by the vertices. We solve for \(a\) by finding the distance between the y -coordinates of the vertices.
\[\begin{align} 2a &=2−(−8) \nonumber \\ 2a &=10 \nonumber\\ a&=5 \nonumber \end{align} \nonumber\]
So \(a^2=25\).
Now we find \(c^2\). The foci are given by \((h,k\pm c)\). So, \((h,k−c)=(−2,−7)\) and \((h,k+c)=(−2,1)\). We substitute \(k=−3\) using either of these points to solve for \(c\).
\[\begin{align} k+c &=1 \nonumber \\ −3+c&=1 \nonumber \\ c&=4 \nonumber \end{align} \nonumber\]
So \(c^2=16\).
Next, we solve for \(b^2\) using the equation \(c^2=a^2−b^2\).
\[\begin{align} c^2&=a^2−b^2 \nonumber \\ 16&=25−b^2 \nonumber \\ b^2&=9 \nonumber \end{align} \nonumber\]
Finally, we substitute the values found for \(h\), \(k\), \(a^2\), and \(b^2\) into the standard form equation for an ellipse:
\[\dfrac{{(x+2)}^2}{9}+\dfrac{{(y+3)}^2}{25}=1 \nonumber\]
What is the standard form equation of the ellipse that has vertices \((−3,3)\) and \((5,3)\) and foci \((1−2\sqrt{3},3)\) and \((1+2\sqrt{3},3)\)?
- Answer
-
\(\dfrac{{(x−1)}^2}{16}+\dfrac{{(y−3)}^2}{4}=1 \nonumber\)
Graphing Ellipses Centered at the Origin
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, a>b\) for horizontal ellipses
and
\(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1, a>b\) for vertical ellipses
-
Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
-
If the equation is in the form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), where \(a>b\), then
- the major axis is the \(x\)-axis
- the coordinates of the vertices are \((\pm a,0)\)
- the coordinates of the co-vertices are \((0,\pm b)\)
- the coordinates of the foci are \((\pm c,0)\)
-
If the equation is in the form \(x^2b^2+y^2a^2=1\),where \(a>b\), then
- the major axis is the \(y\)-axis
- the coordinates of the vertices are \((0,\pm a)\)
- the coordinates of the co-vertices are \((\pm b,0)\)
- the coordinates of the foci are \((0,\pm c)\)
-
If the equation is in the form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), where \(a>b\), then
- Solve for \(c\) using the equation \(c^2=a^2−b^2\).
- Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
Graph the ellipse given by the equation, \(\dfrac{x^2}{9}+\dfrac{y^2}{25}=1\). Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because \(25>9\),the major axis is on the \(y\)-axis. Therefore, the equation is in the form \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\),where \(b^2=9\) and \(a^2=25\). It follows that:
- the center of the ellipse is \((0,0)\)
- the coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{25})=(0,\pm 5)\)
- the coordinates of the co-vertices are \((\pm b,0)=(\pm 9,0)=(\pm 3,0)\)
- the coordinates of the foci are \((0,\pm c)\), where \(c^2=a^2−b^2\) Solving for \(c\), we have:
\[\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\ &=\pm \sqrt{25−9} \nonumber\\ &=\pm \sqrt{16} \nonumber\\ &=\pm 4 \nonumber \end{align} \nonumber\]
Therefore, the coordinates of the foci are \((0,\pm 4)\).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure \(\PageIndex{8}\).
Figure \(\PageIndex{8}\)
Graph the ellipse given by the equation \(\dfrac{x^2}{36}+\dfrac{y^2}{4}=1\). Identify and label the center, vertices, co-vertices, and foci.
- Answer
-
center: \((0,0)\); vertices: \((\pm 6,0)\); co-vertices: \((0,\pm 2)\); foci: \((\pm 4\sqrt{2},0)\)
Figure \(\PageIndex{9}\)
Graph the ellipse given by the equation \(4x^2+25y^2=100\). Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Solution
First, use algebra to rewrite the equation in standard form.
\[\begin{align} 4x^2+25y^2&=100 \nonumber \\ \dfrac{4x^2}{100}+\dfrac{25y^2}{100}&=\dfrac{100}{100} \nonumber \\ \dfrac{x^2}{25}+\dfrac{y^2}{4}&=1 \nonumber \end{align} \nonumber \]
Next, we determine the position of the major axis. Because \(25>4\), the major axis is on the \(x\)-axis. Therefore, the equation is in the form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), where \(a^2=25\) and \(b^2=4\). It follows that:
- the center of the ellipse is \((0,0)\)
- the coordinates of the vertices are \((\pm a,0)=(\pm \sqrt{25},0)=(\pm 5,0)\)
- the coordinates of the co-vertices are \((0,\pm b)=(0,\pm \sqrt{4})=(0,\pm 2)\)
- the coordinates of the foci are \((\pm c,0)\), where \(c^2=a^2−b^2\). Solving for \(c\), we have:
\[\begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\ &=\pm \sqrt{25−4} \nonumber \\ &=\pm \sqrt{21} \nonumber \end{align} \nonumber \]
Therefore the coordinates of the foci are \((\pm \sqrt{21},0)\).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Figure \(\PageIndex{10}\)
Graph the ellipse given by the equation \(49x^2+16y^2=784\). Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
- Answer
-
Standard form: \(\dfrac{x^2}{16}+\dfrac{y^2}{49}=1\); center: \((0,0)\); vertices: \((0,\pm 7)\); co-vertices: \((\pm 4,0)\); foci: \((0,\pm \sqrt{33})\)
Figure \(\PageIndex{11}\)
Graphing Ellipses Not Centered at the Origin
When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, \((h,k)\),we use the standard forms \(\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\), \(a>b\) for horizontal ellipses and \(\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1\), \(a>b\) for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
-
Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
-
If the equation is in the form \(\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\), where \(a>b\), then
- the center is \((h,k)\)
- the major axis is parallel to the \(x\)-axis
- the coordinates of the vertices are \((h\pm a,k)\)
- the coordinates of the co-vertices are \((h,k\pm b)\)
- the coordinates of the foci are \((h\pm c,k)\)
-
If the equation is in the form \(\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1\), where \(a>b\), then
- the center is \((h,k)\)
- the major axis is parallel to the \(y\)-axis
- the coordinates of the vertices are \((h,k\pm a)\)
- the coordinates of the co-vertices are \((h\pm b,k)\)
- the coordinates of the foci are \((h,k\pm c)\)
-
If the equation is in the form \(\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\), where \(a>b\), then
- Solve for \(c\) using the equation \(c^2=a^2−b^2\).
- Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
Graph the ellipse given by the equation, \(\dfrac{{(x+2)}^2}{4}+\dfrac{{(y−5)}^2}{9}=1\). Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because \(9>4\), the major axis is parallel to the \(y\)-axis. Therefore, the equation is in the form \(\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1\), where \(b^2=4\) and \(a^2=9\). It follows that:
- the center of the ellipse is \((h,k)=(−2,5)\)
- the coordinates of the vertices are \((h,k\pm a)=(−2,5\pm \sqrt{9})=(−2,5\pm 3)\), or \((−2,2)\) and \((−2,8)\)
- the coordinates of the co-vertices are \((h\pm b,k)=(−2\pm \sqrt{4},5)=(−2\pm 2,5)\), or \((−4,5)\) and \((0,5)\)
- the coordinates of the foci are \((h,k\pm c)\), where \(c^2=a^2−b^2\). Solving for \(c\),we have:
\[ \begin{align} c&=\pm \sqrt{a^2−b^2} \nonumber \\[4pt] &=\pm \sqrt{9−4} \nonumber \\[4pt] &=\pm \sqrt{5} \nonumber \end{align} \nonumber\]
Therefore, the coordinates of the foci are \((−2,5−\sqrt{5})\) and \((−2,5+\sqrt{5})\).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Figure \(\PageIndex{12}\)
Graph the ellipse given by the equation \(\dfrac{{(x−4)}^2}{36}+\dfrac{{(y−2)}^2}{20}=1\). Identify and label the center, vertices, co-vertices, and foci.
- Answer
-
Center: \((4,2)\); vertices: \((−2,2)\) and \((10,2)\); co-vertices: \((4,2−2\sqrt{5})\) and \((4,2+2\sqrt{5})\); foci: \((0,2)\) and \((8,2)\)
Figure \(\PageIndex{13}\)
- Recognize that an ellipse described by an equation in the form \(ax^2+by^2+cx+dy+e=0\) is in general form.
- Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
- Factor out the coefficients of the \(x^2\) and \(y^2\) terms in preparation for completing the square.
- Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, \(m_1{(x−h)}^2+m_2{(y−k)}^2=m_3\), where \(m_1\), \(m_2\),and \(m_3\) are constants.
- Divide both sides of the equation by the constant term to express the equation in standard form.
Graph the ellipse given by the equation \(4x^2+9y^2−40x+36y+100=0\). Identify and label the center, vertices, co-vertices, and foci.
Solution
We must begin by rewriting the equation in standard form.
\(4x^2+9y^2−40x+36y+100=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
\((4x^2−40x)+(9y^2+36y)=−100\)
Factor out the coefficients of the squared terms.
\(4(x^2−10x)+9(y^2+4y)=−100\)
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
\(4(x^2−10x+25)+9(y^2+4y+4)=−100+100+36\)
Rewrite as perfect squares.
\(4{(x−5)}^2+9{(y+2)}^2=36\)
Divide both sides by the constant term to place the equation in standard form.
\(\dfrac{{(x−5)}^2}{9}+\dfrac{{(y+2)}^2}{4}=1\)
Now that the equation is in standard form, we can determine the position of the major axis. Because \(9>4\), the major axis is parallel to the \(x\)-axis. Therefore, the equation is in the form \(\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\), where \(a^2=9\) and \(b^2=4\). It follows that:
- the center of the ellipse is \((h,k)=(5,−2)\)
- the coordinates of the vertices are \((h\pm a,k)=(5\pm \sqrt{9},−2)=(5\pm 3,−2)\), or \((2,−2)\) and \((8,−2)\)
- the coordinates of the co-vertices are \((h,k\pm b)=(5,−2\pm \sqrt{4})=(5,−2\pm 2)\), or \((5,−4)\) and \((5,0)\)
- the coordinates of the foci are \((h\pm c,k)\), where \(c^2=a^2−b^2\). Solving for \(c\), we have:
\[\begin{align*} c&=\pm \sqrt{a^2-b^2}\\ &=\pm \sqrt{9-4}\\ &=\pm \sqrt{5} \end{align*}\]
Therefore, the coordinates of the foci are \((5−\sqrt{5},−2)\) and \((5+\sqrt{5},−2)\).
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure \(\PageIndex{14}\).
Figure \(\PageIndex{14}\)
Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.
\(4x^2+y^2−24x+2y+21=0\)
- Answer
-
\(\dfrac{{(x−3)}^2}{4}+\dfrac{{(y+1)}^2}{16}=1\); center: \((3,−1)\); vertices: \((3,−5)\) and \((3,3)\); co-vertices: \((1,−1)\) and \((5,−1)\); foci: \((3,−1−2\sqrt{3})\) and \((3,−1+2\sqrt{3})\)
Solving Applied Problems Involving Ellipses
Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus (Figure \(\PageIndex{15}\)). In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about \(43\) feet apart—can hear each other whisper.
Figure \(\PageIndex{15}\): Sound waves are reflected between foci in an elliptical room, called a whispering chamber.
The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are \(46\) feet wide by \(96\) feet long as shown in Figure \(\PageIndex{16}\).
- What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point \((0,0)\).
- If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot.
Figure \(\PageIndex{16}\)
Solution
-
We are assuming a horizontal ellipse with center \((0,0)\), so we need to find an equation of the form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), where \(a>b\). We know that the length of the major axis, \(2a\), is longer than the length of the minor axis, \(2b\). So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.
Therefore, the equation of the ellipse is \[(dfrac{x^2}{2304}+\dfrac{y^2}{529}=1\)
- Solving for \(a\), we have \(2a=96\), so \(a=48\), and \(a^2=2304\).
- Solving for \(b\), we have \(2b=46\), so \(b=23\), and \(b^2=529\).
- To find the distance between the senators, we must find the distance between the foci, \((\pm c,0)\), where \(c^2=a^2−b^2\). Solving for \(c\),we have:
\[\begin{align*} c^2&=a^2-b^2\\ c^2&=2304-529\qquad \text{Substitute using the values found in part } (a)\\ c&=\pm \sqrt{2304-529}\qquad \text{Take the square root of both sides.}\\ c&=\pm \sqrt{1775}\qquad \text{Subtract.}\\ c&\approx \pm 42\qquad \text{Round to the nearest foot.} \end{align*}\]
The points \((\pm 42,0)\) represent the foci. Thus, the distance between the senators is \(2(42)=84\) feet.
Suppose a whispering chamber is \(480\) feet long and \(320\) feet wide.
- What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point \((0,0)\).
- If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.
- Answer a
-
\(\dfrac{x^2}{57,600}+\dfrac{y^2}{25,600}=1\)
- Answer b
-
The people are standing \(358\) feet apart.
Access these online resources for additional instruction and practice with ellipses.
Key Equations
| Horizontal ellipse, center at origin | \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), \(a>b\) |
| Vertical ellipse, center at origin | \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\), \(a>b\) |
| Horizontal ellipse, center \((h,k)\) | \(\dfrac{{(x−h)}^2}{a^2}+\dfrac{{(y−k)}^2}{b^2}=1\), \(a>b\) |
| Vertical ellipse, center \((h,k)\) | \(\dfrac{{(x−h)}^2}{b^2}+\dfrac{{(y−k)}^2}{a^2}=1\), \(a>b\) |
Key Concepts
- An ellipse is the set of all points \((x,y)\) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
- When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See Example \(\PageIndex{1}\) and Example \(\PageIndex{2}\).
- When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See Example \(\PageIndex{3}\) and Example \(\PageIndex{4}\).
- When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See Example \(\PageIndex{5}\) and Example \(\PageIndex{6}\).
- Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See Example \(\PageIndex{7}\).
Contributors and Attributions
-
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .


1.1: Earth and Sky
Learning Objectives
By the end of this section, you will be able to:
- Describe how latitude and longitude are used to map Earth
- Explain how right ascension and declination are used to map the sky
In order to create an accurate map, a mapmaker needs a way to uniquely and simply identify the location of all the major features on the map, such as cities or natural landmarks. Similarly, astronomical mapmakers need a way to uniquely and simply identify the location of stars, galaxies, and other celestial objects. On Earth maps, we divide the surface of Earth into a grid, and each location on that grid can easily be found using its latitude and longitude coordinate. Astronomers have a similar system for objects on the sky. Learning about these can help us understand the apparent motion of objects in the sky from various places on Earth.
Locating Places on Earth
Let’s begin by fixing our position on the surface of planet Earth. Earth’s axis of rotation defines the locations of its North and South Poles and of its equator halfway between. Two other directions are also defined by Earth’s motions: east is the direction toward which Earth rotates, and west is its opposite. At almost any point on Earth, the four directions—north, south, east, and west—are well defined, despite the fact that our planet is round rather that flat. The only exceptions are exactly at the North and South Poles, where the directions east and west are ambiguous (because points exactly at the poles do not turn).
We can use these ideas to define a system of coordinates attached to our planet. Such a system, like the layout of streets and avenues in Manhattan or Salt Lake City, helps us find where we are or want to go. Coordinates on a sphere, however, are a little more complicated than those on a flat surface. We must define circles on the sphere that play the same role as the rectangular grid that you see on city maps.
A great circle is any circle on the surface of a sphere whose center is at the center of the sphere. For example, Earth’s equator is a great circle on Earth’s surface, halfway between the North and South Poles. We can also imagine a series of great circles that pass through both the North and South Poles. Each of the circles is called a meridian ; they are each perpendicular to the equator, crossing it at right angles.
Any point on the surface of Earth will have a meridian passing through it (Figure \(\PageIndex{1}\)). The meridian specifies the east-west location, or longitude, of the place. By international agreement (and it took many meetings for the world’s countries to agree), longitude is defined as the number of degrees of arc along the equator between your meridian and the one passing through Greenwich, England, which has been designated as the Prime Meridian. The longitude of the Prime Meridian is defined as 0°.
Why Greenwich, you might ask? Every country wanted 0° longitude to pass through its own capital. Greenwich, the site of the old Royal Observatory (Figure \(\PageIndex{2}\)), was selected because it was between continental Europe and the United States, and because it was the site for much of the development of the method to measure longitude at sea. Longitudes are measured either to the east or to the west of the Greenwich meridian from 0° to 180°. As an example, the longitude of the clock-house benchmark of the U.S. Naval Observatory in Washington, DC, is 77.066° W.
Your latitude (or north-south location) is the number of degrees of arc you are away from the equator along your meridian. Latitudes are measured either north or south of the equator from 0° to 90°. (The latitude of the equator is 0°.) As an example, the latitude of the previously mentioned Naval Observatory benchmark is 38.921° N. The latitude of the South Pole is 90° S, and the latitude of the North Pole is 90° N.
Locating Places in the Sky
Positions in the sky are measured in a way that is very similar to the way we measure positions on the surface of Earth. Instead of latitude and longitude, however, astronomers use coordinates called declination and right ascension . To denote positions of objects in the sky, it is often convenient to make use of the fictitious celestial sphere. The sky appears to rotate about points above the North and South Poles of Earth—points in the sky called the north celestial pole and the south celestial pole . Halfway between the celestial poles, and thus 90° from each pole, is the celestial equator , a great circle on the celestial sphere that is in the same plane as Earth’s equator. We can use these markers in the sky to set up a system of celestial coordinates.
Declination on the celestial sphere is measured the same way that latitude is measured on the sphere of Earth: from the celestial equator toward the north (positive) or south (negative). So Polaris, the star near the north celestial pole, has a declination of almost +90°.
Right ascension (RA) is like longitude, except that instead of Greenwich, the arbitrarily chosen point where we start counting is the vernal equinox , a point in the sky where the ecliptic (the Sun’s path) crosses the celestial equator. RA is expressed in units of time and ranges from 0 hours to 23 hours 59 minutes 59 seconds. The celestial sphere appears to turn around Earth once a day as our planet turns on its axis. Since the Earth is a sphere, it takes 24 hours to rotate 360° meaning that every 15° is equal to 1 hour of time. For example, the approximate celestial coordinates of the bright star Capella are RA 5h and declination +50°.
One way to visualize these circles in the sky is to imagine Earth as a transparent sphere with the terrestrial coordinates (latitude and longitude) painted on it with dark paint. Imagine the celestial sphere around us as a giant ball, painted white on the inside. Then imagine yourself at the center of Earth, with a bright light bulb in the middle, looking out through its transparent surface to the sky. The terrestrial poles, equator, and meridians will be projected as dark shadows on the celestial sphere, giving us the system of coordinates in the sky.
You can explore a variety of basic animations about coordinates and motions in the sky at this interactive site from ClassAction. Click on the “Animations” tab for a list of options. If you choose the second option in the menu, you can play with the celestial sphere and see RA and declination defined visually.
The Turning Earth
Why do many stars rise and set each night? Why, in other words, does the night sky seem to turn? We have seen that the apparent rotation of the celestial sphere could be accounted for either by a daily rotation of the sky around a stationary Earth or by the rotation of Earth itself. Since the seventeenth century, it has been generally accepted that it is Earth that turns, but not until the nineteenth century did the French physicist Jean Foucault provide an unambiguous demonstration of this rotation. In 1851, he suspended a 60-meter pendulum weighing about 25 kilograms from the dome of the Pantheon in Paris and started the pendulum swinging evenly. If Earth had not been turning, there would have been no alteration of the pendulum’s plane of oscillation, and so it would have continued tracing the same path. Yet after a few minutes Foucault could see that the pendulum’s plane of motion was turning. Foucault explained that it was not the pendulum that was shifting, but rather Earth that was turning beneath it (Figure \(\PageIndex{3}\)). You can now find such pendulums in many science centers and planetariums around the world.
Can you think of other pieces of evidence that indicate that it is Earth and not the sky that is turning? (See Section 4.E, Collaborative Group Activity A at the end of this chapter.)
Key Concepts and Summary
The terrestrial system of latitude and longitude makes use of the great circles called meridians. Longitude is arbitrarily set to 0° at the Royal Observatory at Greenwich, England. An analogous celestial coordinate system is called right ascension (RA) and declination, with 0° of declination at the equator and 0 hours of RA starting at the vernal equinox. These coordinate systems help us locate any object on the celestial sphere. The Foucault pendulum is a way to demonstrate that Earth is turning.
Glossary
- declination
- the angular distance north or south of the celestial equator
- great circle
- a circle on the surface of a sphere that is the curve of intersection of the sphere with a plane passing through its center
- meridian
- a great circle on the terrestrial or celestial sphere that passes through the poles
- right ascension
- the coordinate for measuring the east-west positions of celestial bodies; the angle measured eastward along the celestial equator from the vernal equinox to the hour circle passing through a body
Contributors
-
Andrew Fraknoi (Foothill College), David Morrison (NASA Ames Research Center), Sidney C. Wolff (National Optical Astronomy Observatory) with many contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/astronomy ).


6.7: Exercise, Nutrition, Hormones, and Bone Tissue
By the end of this section, you will be able to:
- Describe the effect exercise has on bone tissue
- List the nutrients that affect bone health
- Discuss the role those nutrients play in bone health
- Describe the effects of hormones on bone tissue
All of the organ systems of your body are interdependent, and the skeletal system is no exception. The food you take in via your digestive system and the hormones secreted by your endocrine system affect your bones.
Exercise and Bone Tissue
During long space missions, astronauts can lose approximately 1 to 2 percent of their bone mass per month. This loss of bone mass is thought to be caused by the lack of mechanical stress on astronauts’ bones due to the low gravitational forces in space. Lack of mechanical stress causes bones to lose mineral salts and collagen fibers, and thus strength. Similarly, mechanical stress stimulates the deposition of mineral salts and collagen fibers. The internal and external structure of a bone will change as stress increases or decreases so that the bone is an ideal size and weight for the amount of activity it endures. That is why people who exercise regularly have thicker bones than people who are more sedentary. It is also why a broken bone in a cast atrophies while its contralateral mate maintains its concentration of mineral salts and collagen fibers. The bones undergo remodeling as a result of forces (or lack of forces) placed on them.
Numerous, controlled studies have demonstrated that people who exercise regularly have greater bone density than those who are more sedentary. Any type of exercise will stimulate the deposition of more bone tissue, but resistance training has a greater effect than cardiovascular activities. Resistance training is especially important to slow down the eventual bone loss due to aging and for preventing osteoporosis.
Nutrition and Bone Tissue
The vitamins and minerals contained in all of the food we consume are important for all of our organ systems. However, there are certain nutrients that affect bone health.
Calcium and Vitamin D
You already know that calcium is a critical component of bone, especially in the form of calcium phosphate and calcium carbonate. Since the body cannot make calcium, it must be obtained from the diet. However, calcium cannot be absorbed from the small intestine without vitamin D. Therefore, intake of vitamin D is also critical to bone health. In addition to vitamin D’s role in calcium absorption, it also plays a role, though not as clearly understood, in bone remodeling.
Milk and other dairy foods are not the only sources of calcium. This important nutrient is also found in green leafy vegetables, broccoli, and intact salmon and canned sardines with their soft bones. Nuts, beans, seeds, and shellfish provide calcium in smaller quantities.
Except for fatty fish like salmon and tuna, or fortified milk or cereal, vitamin D is not found naturally in many foods. The action of sunlight on the skin triggers the body to produce its own vitamin D (Figure \(\PageIndex{1}\)), but many people, especially those of darker complexion and those living in northern latitudes where the sun’s rays are not as strong, are deficient in vitamin D. In cases of deficiency, a doctor can prescribe a vitamin D supplement.
Other Nutrients
Vitamin K also supports bone mineralization and may have a synergistic role with vitamin D in the regulation of bone growth. Green leafy vegetables are a good source of vitamin K.
The minerals magnesium and fluoride may also play a role in supporting bone health. While magnesium is only found in trace amounts in the human body, more than 60 percent of it is in the skeleton, suggesting it plays a role in the structure of bone. Fluoride can displace the hydroxyl group in bone’s hydroxyapatite crystals and form fluorapatite. Similar to its effect on dental enamel, fluorapatite helps stabilize and strengthen bone mineral. Fluoride can also enter spaces within hydroxyapatite crystals, thus increasing their density.
Omega-3 fatty acids have long been known to reduce inflammation in various parts of the body. Inflammation can interfere with the function of osteoblasts, so consuming omega-3 fatty acids, in the diet or in supplements, may also help enhance production of new osseous tissue. Table 6.5 summarizes the role of nutrients in bone health.
| Nutrient | Role in bone health |
|---|---|
| Calcium | Needed to make calcium phosphate and calcium carbonate, which form the hydroxyapatite crystals that give bone its hardness |
| Vitamin D | Needed for calcium absorption |
| Vitamin K | Supports bone mineralization; may have synergistic effect with vitamin D |
| Magnesium | Structural component of bone |
| Fluoride | Structural component of bone |
| Omega-3 fatty acids | Reduces inflammation that may interfere with osteoblast function |
Hormones and Bone Tissue
The endocrine system produces and secretes hormones, many of which interact with the skeletal system. These hormones are involved in controlling bone growth, maintaining bone once it is formed, and remodeling it.
Hormones That Influence Osteoblasts and/or Maintain the Matrix
Several hormones are necessary for controlling bone growth and maintaining the bone matrix. The pituitary gland secretes growth hormone (GH), which, as its name implies, controls bone growth in several ways. It triggers chondrocyte proliferation in epiphyseal plates, resulting in the increasing length of long bones. GH also increases calcium retention, which enhances mineralization, and stimulates osteoblastic activity, which improves bone density.
GH is not alone in stimulating bone growth and maintaining osseous tissue. Thyroxine, a hormone secreted by the thyroid gland promotes osteoblastic activity and the synthesis of bone matrix. During puberty, the sex hormones (estrogen and testosterone) also come into play. They too promote osteoblastic activity and production of bone matrix, and in addition, are responsible for the growth spurt that often occurs during adolescence. They also promote the conversion of the epiphyseal plate to the epiphyseal line (i.e., cartilage to its bony remnant), thus bringing an end to the longitudinal growth of bones. Additionally, calcitriol, the active form of vitamin D, is produced by the kidneys and stimulates the absorption of calcium and phosphate from the digestive tract.
Aging and the...
Skeletal System
Osteoporosis is a disease characterized by a decrease in bone mass that occurs when the rate of bone resorption exceeds the rate of bone formation, a common occurrence as the body ages. Notice how this is different from Paget’s disease. In Paget’s disease, new bone is formed in an attempt to keep up with the resorption by the overactive osteoclasts, but that new bone is produced haphazardly. In fact, when a physician is evaluating a patient with thinning bone, they will test for osteoporosis and Paget’s disease (as well as other diseases). Osteoporosis does not have the elevated blood levels of alkaline phosphatase found in Paget’s disease.
While osteoporosis can involve any bone, it most commonly affects the proximal ends of the femur, vertebrae, and wrist. As a result of the loss of bone density, the osseous tissue may not provide adequate support for everyday functions, and something as simple as a sneeze can cause a vertebral fracture. When an elderly person falls and breaks a hip (really, the femur), it is very likely the femur that broke first, which resulted in the fall. Histologically, osteoporosis is characterized by a reduction in the thickness of compact bone and the number and size of trabeculae in cancellous bone.
Figure \(\PageIndex{2}\) shows that females lose bone mass more quickly than males starting at about 50 years of age. This occurs because 50 is the approximate age at which females go through menopause. Not only do their menstrual periods lessen and eventually cease, but their ovaries reduce in size and then cease the production of estrogen, a hormone that promotes osteoblastic activity and production of bone matrix. Thus, osteoporosis is more common in females, but males can develop it, too. Anyone with a family history of osteoporosis has a greater risk of developing the disease, so the best treatment is prevention, which should start with a childhood diet that includes adequate intake of calcium and vitamin D and a lifestyle that includes weight-bearing exercise. These actions, as discussed above, are important in building bone mass. Promoting proper nutrition and weight-bearing exercise early in life can maximize bone mass before the age of 30, thus reducing the risk of osteoporosis.
For many elderly people, a hip fracture can be life threatening. The fracture itself may not be serious, but the immobility that comes during the healing process can lead to the formation of blood clots that can lodge in the capillaries of the lungs, resulting in respiratory failure; pneumonia due to the lack of poor air exchange that accompanies immobility; pressure sores (bed sores) that allow pathogens to enter the body and cause infections; and urinary tract infections from catheterization.
Current treatments for managing osteoporosis include bisphosphonates (the same medications often used in Paget’s disease), calcitonin, and estrogen (for females only). Minimizing the risk of falls, for example, by removing tripping hazards, is also an important step in managing the potential outcomes from the disease.
Hormones That Influence Osteoclasts
Bone modeling and remodeling require osteoclasts to resorb unneeded, damaged, or old bone, and osteoblasts to lay down new bone. Two hormones that affect the osteoclasts are parathyroid hormone (PTH) and calcitonin.
PTH stimulates osteoclast proliferation and activity. As a result, calcium is released from the bones into the circulation, thus increasing the calcium ion concentration in the blood. PTH also promotes the reabsorption of calcium by the kidney tubules, which can affect calcium homeostasis (see below).
The small intestine is also affected by PTH, albeit indirectly. Because another function of PTH is to stimulate the synthesis of vitamin D, and because vitamin D promotes intestinal absorption of calcium, PTH indirectly increases calcium uptake by the small intestine. Calcitonin, a hormone secreted by the thyroid gland, has some effects that counteract those of PTH. Calcitonin inhibits osteoclast activity and stimulates calcium uptake by the bones, thus reducing the concentration of calcium ions in the blood. As evidenced by their opposing functions in maintaining calcium homeostasis, PTH and calcitonin are generally not secreted at the same time. Table \(\PageIndex{2}\) summarizes the hormones that influence the skeletal system.
| Hormone | Role |
|---|---|
| Growth hormone | Increases length of long bones, enhances mineralization, and improves bone density |
| Thyroxine | Stimulates bone growth and promotes synthesis of bone matrix |
| Sex hormones | Promote osteoblastic activity and production of bone matrix; responsible for adolescent growth spurt; promote conversion of epiphyseal plate to epiphyseal line |
| Calcitriol | Stimulates absorption of calcium and phosphate from digestive tract |
| Parathyroid hormone | Stimulates osteoclast proliferation and resorption of bone by osteoclasts; promotes reabsorption of calcium by kidney tubules; indirectly increases calcium absorption by small intestine |
| Calcitonin | Inhibits osteoclast activity and stimulates calcium uptake by bones |


8.1: Why It Matters- Cuando te gusta tu trabajo, todos los días son vacaciones
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El Bebeto –
Que Te Duela Más Que A Mí
El Bebeto is an award-winning singer of Mexican Regional music from Guasave, Sinaloa.
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Ivy Queen –
Menor que yo
Ivy Queen, is a Puerto Rican singer, songwriter, rapper, actress and record producer. She was born in Añasco, Puerto Rico, but later moved with her parents to New York City. She is often referred to as the Queen of Reggaeton. (
Wikipedia entry on Ivy Queen
)
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Daddy Yankee –
El Mejor De Todos Los Tiempos
Ramón Luis Ayala Rodríguez, known professionally as Daddy Yankee, is a Puerto Rican singer, songwriter, rapper, actor and record producer.
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Melendi –
Tan Tonto Como Tú
Ramón Melendi Espina, known as Melendi, is a Spanish pop singer-songwriter
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7.2: Application of Eigenfunction Series
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The eigenfunction series can arise even from higher order equations. Consider an elastic beam (say made of steel). We will study the transversal vibrations of the beam. That is, suppose the beam lies along the \(x\)-axis and let \(y(x,t)\) measure the displacement of the point \(x\) on the beam at time \(t\). See Figure \(\PageIndex{1}\).
The equation that governs this setup is
\[ a^4 \frac{\partial^4y}{\partial x^4}+\frac{\partial^2y}{\partial t^2}=0, \nonumber \]
for some constant \(a>0\), let us not worry about the physics \(^{1}\) .
Suppose the beam is of length \(1\) simply supported (hinged) at the ends. The beam is displaced by some function \(f(x)\) at time \(t=0\) and then let go (initial velocity is \(0\)). Then \(y\) satisfies:
\[ \begin{align} & a^4 y_{xxxx} + y_{tt} = 0 \qquad (0 < x < 1, \enspace t > 0),\nonumber \\ \label{eq:2} & y(0,t) = y_{xx}(0,t) = 0 , \\ & y(1,t) = y_{xx}(1,t) = 0 , \\ & y(x,0) = f(x), \qquad y_{t}(x,0) = 0 .\nonumber \end{align} \nonumber \]
Again we try \(y(x,t)=X(x)T(t)\) and plug in to get \(a^4X^{(4)}T+XT''=0\) or
\[\frac{X^{(4)}}{X}=\frac{-T''}{a^4T}=\lambda . \nonumber \]
The equations are \[T'' + \lambda a^4 T = 0, \qquad X^{(4)} - \lambda X = 0 . \nonumber \] The boundary conditions \(y(0,t) = y_{xx}(0,t) = 0\) and \(y(1,t) = y_{xx}(1,t) = 0\) imply \[X(0) = X''(0) = 0, \qquad \text{and} \qquad X(1) = X''(1) = 0 . \nonumber \]
The initial homogeneous condition \(y_t(x,0) = 0\) implies \[T'(0) = 0 . \nonumber \] As usual, we leave the nonhomogeneous \(y(x,0) = f(x)\) for later.
Considering the equation for \(T\), that is, \(T'' + \lambda a^4 T = 0\), and physical intuition leads us to the fact that if \(\lambda\) is an eigenvalue then \(\lambda > 0\): We expect vibration and not exponential growth nor decay in the \(t\) direction (there is no friction in our model for instance). So there are no negative eigenvalues. Similarly \(\lambda = 0\) is not an eigenvalue.
Exercise \(\PageIndex{1}\)
Justify \(\lambda >0\) just from the equation for \(X\) and the boundary conditions.
Write \( \omega^4=\lambda\), so that we do not need to write the fourth root all the time. For \(X\) we get the equation \(X^{(4)}- \omega^4X=0\). The general solution is
\[ X(x)=Ae^{\omega x}+Be^{- \omega x}+C\sin(\omega x)+D\cos(\omega x). \nonumber \]
Now \(0=X(0)A+B+D, 0=X''(0)=\omega^2(A+B-D)\). Hence, \(D=0\) and \(A+B=0\), or \(B=-A\). So we have
\[ X(x)=Ae^{\omega x}-Ae^{- \omega x}+C\sin(\omega x). \nonumber \]
Also \(0=X(1)=A(e^{\omega}-e^{- \omega})+C\sin \omega\), and \(0=X''(1)=A\omega^2(e^{\omega}-e^{- \omega})-C\omega^2\sin \omega\). This means that \(C\sin \omega =0\) and \(A(e^{\omega}-e^{- \omega})=2A\sinh \omega=0\). If \(\omega >0\), then \(\omega \neq 0\) and so \(A=0\). This means that \(C \neq 0\) otherwise \(\lambda\) is not an eigenvalue. Also \(\omega\) must be an integer multiple of \(\pi\). Hence \(\omega =n\pi\) and \(n \geq 1\) (as \(\omega >0\)). We can take \(C=1\). So the eigenvalues are \(\lambda_n=n^4\pi^4\) and the eigenfunctions are \(\sin(n\pi x)\).
Now \(T''+n^4\pi^4a^4T=0\). The general solution is \( T(t)=A\sin(n^2\pi^2a^2t)+B\cos(n^2\pi^2a^2t)\). But \(T'(0)=0\) and hence we must have \(A=0\) and we can take \(B=1\) to make \(T(0)=1\) for convenience. So our solutions are \( T_n(t)=\cos(n^2\pi^2a^2t)\).
As the eigenfunctions are just sines again, we can decompose the function \(f(x)\) on \(0<x<1\) using the sine series. We find numbers \(b_n\) such that for \(0<x<1\) we have
\[ f(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x). \nonumber \]
Then the solution to \(\eqref{eq:2}\) is
\[ y(x,t)=\sum_{n=1}^{\infty}b_nX_n(x)T_n(t)=\sum_{n=1}^{\infty}b_n\sin(n\pi x)\cos(n^2\pi^2a^2t). \nonumber \]
The point is that \(X_nT_n\) is a solution that satisfies all the homogeneous conditions (that is, all conditions except the initial position). And since and \(T_n(0)=1\), we have
\[ y(x,0)=\sum_{n=1}^{\infty}b_nX_n(x)T_n(0)=\sum_{n=1}^{\infty}b_nX_n(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x)=f(x). \nonumber \]
So \(y(x,t)\) solves \(\eqref{eq:2}\).
The natural (circular) frequencies of the system are \(n^2\pi^2a^2\). These frequencies are all integer multiples of the fundamental frequency \(\pi^2a^2\), so we get a nice musical note. The exact frequencies and their amplitude are what we call the timbre of the note.
The timbre of a beam is different than for a vibrating string where we get “more” of the lower frequencies since we get all integer multiples, \(1,2,3,4,5, \ldots\). For a steel beam we get only the square multiples \(1,4,9,16,25, \ldots\). That is why when you hit a steel beam you hear a very pure sound. The sound of a xylophone or vibraphone is, therefore, very different from a guitar or piano.
Example \(\PageIndex{1}\)
Let us assume that \(f(x)= \frac{x(x-1)}{10}\). On \(0<x<1\) we have (you know how to do this by now)
\[ f(x)= \sum_{ \underset{\rm{n~odd}}{n=1}}^{\infty}\frac{4}{5\pi^3n^3}\sin(n\pi x). \nonumber \]
Hence, the solution to \(\eqref{eq:2}\) with the given initial position \(f(x)\) is
\[ y(x,t)= \sum_{ \underset{\rm{n~odd}}{n=1}}^{\infty}\frac{4}{5\pi^3n^3}\sin(n\pi x)\cos(n^2\pi^2a^2t). \nonumber \]
There are other boundary conditions than just hinged ends. There are three basic possibilities: hinged, free, or fixed. Let us consider the end at \(x=0\). For the other end, it is the same idea. If the end is hinged , then \[u(0,t) = u_{xx}(0,t) = 0 . \nonumber \] If the end is free , that is, it is just floating in air, then \[u_{xx}(0,t) = u_{xxx}(0,t) = 0 . \nonumber \] And finally, if the end is clamped or fixed , for example it is welded to a wall, then \[u(0,t) = u_{x}(0,t) = 0 . \nonumber \]
Footnotes
[1] If you are interested, \(a^{4}=\frac{EI}{\rho}\), where \(E\) is the elastic modulus, \(I\) is the second moment of area of the cross section, and \(\rho\) is linear density.