<s>[INST] Trying to find a general equation for an ellipse given the foci and sum of focal distances I'm trying to find an equation for the ellipse in the form $$Ax^2 + Bxy + Cy^2 +Dx +Ey +F = 0$$ given the foci $(a,b)$ and $(c,d)$ and the sum of focal distances $r$. I started from the definition $$\sqrt{(x-a)^2+(y-b)^2} + \sqrt{(x-c)^2+(y-d)^2} = r$$ squared both sides, moved not radical terms to the right side and squared again and after trudging through a lot of algebra I've arrived at the equation (easier reading of coefficients below)
\begin{align}
0 &= (r^2 + (a-c)^2)x^2 + 2(a-c)(b-d)xy + (r^2 + (b-d)^2)y^2\\
&\qquad + (r^2(a+c) - (a-c)(a^2+b^2-c^2-d^2))x \\
&\qquad +(r^2(b+d) - (b-d)(a^2+b^2-c^2-d^2))y \\
&\qquad+ \frac{1}{4}(r^4 + 2r^2(a^2+b^2+c^2+d^2) + (a^2+b^2-c^2-d^2)^2),
\end{align}
which I want to believe is close, but this does not produce a graph on Desmos.  If anyone just has a reference for the equation that I might be able to look at and find my mistakes that would be much appreciated. When I looked on Wikipedia, they talked about using the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 $$ and rotating the major axis, but I have no idea how to translate those coefficients to be in terms of the foci. For easier reading my coefficients are
\begin{align}
A &= r^2+(a-c)^2 \\
B &= 2(a-c)(b-d) \\
C &= r^2 + (b-d)^2 \\
D &= r^2(a+c)-(a-c)(a^2+b^2-c^2-d^2) \\
E &= r^2(b+d)-(b-d)(a^2+b^2-c^2-d^2) \\
F &= \frac{1}{4}(r^4 +2r^2(a^2+b^2+c^2+d^2) + (a^2+b^2-c^2-d^2)^2).
\end{align}
If the middle term of F was $2r^2(a^2+b^2-c^2-d^2)$ I could factor it, but because the left hand side (after the second squaring round) has no $r$ term, there isn't an opportunity for it to change like all the other terms did. Sorry I can't be more specific, but I didn't think typing out the mountain of algebra I've done was a good idea.
 [/INST] Doing it for $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$ is fun.
$S_1(-ae,0(, S_2(ae,0)$,  $P(x,y)$ given that 
$$PS_1+PS_2=2a \implies \sqrt{(x+ae)^2+y^2}+\sqrt{(x-ae)^2+y^2}=2a$$ $$\implies U+V=2a~~~~~(1)$$
Then $$U^2-V^2=4aex ~~~ U-V=2ex~~~~~~(3)$$
From (1) and (2), we get
$$U=a+ex \implies (x+ae)^2+y^2=(a+ex)^2$$
$$\implies x^2+y^2-e^2x^2=a^2-a^2e^2 \implies (1-e^2) x^2+y^2=b^2$$
$$\implies \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
</s>