<s>[INST] Finding equation of hyperbola with only foci and asymptote This is a concept we learned in class today, which I still can't seem to grasp. I have no specific question that necessarily has to be done, so I will use one of the examples my book gives me:
Given the foci are: (6√5, 10) and (-6√5,10) 
The asymptote is y = 1/2x + 10 
Find the equation of the hyperbola
I know that...
(y-k)²/a² - (x-h)²/b² = 1 (Hyperbola equation)
Center (h,k)
foci (h, k±c), c² = a²+b² 
Asmyptote y = -(a/b)x + k+(a/b)h (The -/+ are interchangeable) 
I am just confused on how to find the equation with the givens above. Thanks!
 [/INST] It looks like you know all of the equations you need to solve this problem.  I also see that you know that the slope of the asymptote line of a hyperbola is the ratio $\dfrac{b}{a}$ for a simple hyperbola of the form $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$
So, given $y=\dfrac{1}{2}x+10$, we can see that the ratio $\dfrac{b}{a}=\dfrac{1}{2}$.  Furthermore, we can say that $b=n$ and $a=2n$ since the $n's$ would cancel in the ratio.
Now that we have our $a$ and $b$ depending on only one variable $n$, we can use the relationship you mentioned, $c^2=a^2+b^2$, to determine the value of $n$.
\begin{align*}
a^2+b^2&=c^2\\
(2n)^2+n^2&=(6\sqrt{5})^2\\
n^2&=36\\
n&=6\\
\end{align*}
Therefore, we have $a=12$ and $b=6$, so then our equation for this hyperbola can be expressed with the following:
\begin{align*}
\dfrac{x^2}{144}-\dfrac{(y-10)^2}{36}=1
\end{align*}
Where the 10 came from shifting the hyperbola up 10 units to match the $y$ value of our foci.
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