<s>[INST] Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$ Yes, I know this is the equation of an ellipse. 
This is the starting point of the problem:
$$
r_1+r_2=2a,
$$
where
$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$
I want to establish certain properties prior to deriving a parametric expression for the ellipse, or even the "standard form",
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
$$
I can do all of this with a couple tacks, a string and a pencil, but
I want to show things algebraically. Preferably in a way that is directly applicable to all second-degree plane curves.
The first property I wish to establish is the range of $x$ values.
I know from experience that the answer is going to be that the extreme
values of $x$ will occur when $y=0$. The question that I am asking
in this post is: what is the most direct (algebraic) way to show that $\left|x\right|$
is greatest when $y=0$? 
I had hoped to establish that fact, and then use another (absolute value)
method of determining the actual values of $x$ when $y=0$. But, the best
I cam up with is this very long-winded demonstration.
First some definitions: Let $a,c,x,y\in\mathbb{R}$ where $a>c>0$
are constants, and $x$ and $y$ are variables such that, for $r_1\equiv\sqrt{(x+c)^2+y^2}$
and $r_2\equiv\sqrt{(x-c)^2+y^2}$ we have the equation
$$
r_1+r_2=2a.
$$
The degenerate case has been omitted.
I will presume the greatest value to be $\left|x\right|=a$, and take
the case of $x=a$. We square our equation
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(a+c)^2+y^2+2r_1 r_2+(a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2.
$$
So
$$
r_1(2a-r_1)=a^2-c^2-y^2
$$
$$
=2a\sqrt{(a+c)^2+y^2}-((a+c)^2+y^2)=a^2-c^2-y^2.
$$
Transpose, cancel and simplify
$$
2a\sqrt{(a+c)^2+y^2}=2a^2+2ac,
$$
$$
r_1=\sqrt{(a+c)^2+y^2}=a+c.
$$
And, therefore
$$
r_2=2a-r_1=a-c.
$$
Now, by definition we have
$$
\sqrt{\left(a+c\right)^2}\equiv\left|a+c\right|=r_1
$$
and
$$
\sqrt{(a-c)^2}\equiv\left|a-c\right|=r_2.
$$
If $y\ne0$ while $x=a$, then 
$$
r_1=\sqrt{(a+c)^2+y^2}=\left|a+c\right|+\delta_1,
$$
and 
$$
r_{2}=\sqrt{(a-c)^2+y^2}=\left|a-c\right|+\delta_2,
$$
where both $\delta_1>0$ and $\delta_2>0.$
Since $a>c>0,$ we have $r_1+r_2=2a+\delta_1+\delta_2$, in
contradiction to the equation we seek to satisfy. So $x=a\implies y=0.$ 
If $x>a$, let $\delta_3=x-a>0.$ 
So 
$$
r_1+r_2=a+c+\delta_3+a-c+\delta_3>2a;
$$
another contradiction.
Thus far we have dertermined that $x=a\implies y=0,$ and for any
$y$ that $x\le a$.
If we set $x=-a,$ then
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(-a+c)^2+y^2+2r_1 r_2+(-a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2
$$
$$
r_{1}\left(2a-r_{1}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}-\left(\left(-a+c\right)^{2}+y^{2}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}=2a^{2}-2ac
$$
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c>0.
$$
$$
r_{2}=2a-r_{1}=a+c.
$$
Since we now have 
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c,
$$
and
$$
r_{2}=\sqrt{\left(a+c\right)^{2}+y^{2}}=a+c,
$$
the argument showing that $x=-a\implies y=0,$ is essentially the
same. Likewise for the argument that $x\ge-a.$
And that's just the $x$ range.
NB: I don't "need" any of this to fully characterize the ellipse.  My goal is to find multiple ways of examining the same construct.
 [/INST] This is a constrained optimization problem that can be attacked directly with the method of Lagrange multipliers. Form the Lagrangian $x-\lambda(r_1+r_2-2a)$ and differentiate, producing the equations $$1-\lambda\left({x+c \over r_1}+{x-c \over r_2}\right) = 0 \\
-\lambda\left({y\over r_1}+{y\over r_2}\right) = 0.$$ The multiplier $\lambda$ cannot be zero and the two distances are always positive, so we must have $y=0$. Substituting this back into the equation $r_1+r_2=2a$ yields $x=\pm a$.  
Strictly speaking, you should also show that $|x|$ can’t be greater than $a$, but that’s almost trivial.  

The above is one of the most straightforward solutions that I can think of, but if you insist on avoiding calculus, here’s another approach. The perimeter of the triangle formed by the foci and an arbitrary point on the ellipse is fixed at $r_1+r_2+2c=2(a+c)$. Using Heron's formula and the defining equation $r_1+r_2=2a$, the square of this triangle’s area can be expressed as a function of $r_1$: $$A^2 = (c-a+r_1)(c+a-r_1)(a+c)(a-c).\tag1$$ Since $a\gt c$, for this to be nonnegative we must have $(c-a+r_1)(c+a-r_1)\ge0$, or $$(r_1-a)^2\le c^2.\tag2$$ On the other hand, by construction every point on the ellipse is also the intersection of the circles $(x+c)^2+y^2=r_1^2$ and $(x-c)^2+y^2=r_2^2=(2a-r_1)^2$. From these equations we get $r_1=\frac ca x+a$. Substituting into (2) and simplifying results in $|x|\le a$. Plugging $\pm a$ into the defining equation of the ellipse and solving for $y$ yields $y=0$, as desired.  
The bounds for $y$ can be found in a similar way. The $y$-coordinate of a point on the ellipse is the altitude of the triangle, so the maximum $y$ value will maximize the triangle’s area. Equation (1) is quadratic in $r_1$ and the $r_1^2$ term has a negative coefficient, so the value of $r_1$ that maximizes the area can be computed using standard algebraic techniques, and from that the corresponding values of $y$.
That wasn’t too bad, but I don’t really see much advantage to working directly with the radicals from the equation $r_1+r_2=2a$. It’s only a couple of more steps from $ar_1=cx+a^2$ to the standard equation of an ellipse. Once you have that equation, the ranges for $x$ and $y$ can be found by inspection, as Michael Hardy wrote in his now-deleted answer.
</s>