Ellipse formula

Question

In the video https://youtu.be/eI4an8aSsgw?t=16354 the professor says that the following equation $$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a$$ simplifies to $$ (a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2) $$ I don't get it. I need someone who can explain it step by step

Answer 1

Let $$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a=\sqrt{P}+\sqrt{Q}~~~(1)$$ $$\implies P-Q=4cx \implies \sqrt{P}-\sqrt{Q}=2cx/a~~~(2)$$ Adding (1) and (2) get $$\sqrt{P}=xc/a+a \implies (x+c)^2+y^2=(cx/a+a)^2 \implies (1-c^2/a^2)x^2+y^2=a^2-c^2$$ $$\implies \frac{x^2}{a^2}+\frac{y^2}{(1-c^2/a^2)}=1$$ Let $c=ae$, then we get ellips as $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$

Answer 2

Let's call $\begin{cases}U=\sqrt{(x+c)^2+y^2}\\V=\sqrt{(x-c)^2+y^2}\end{cases}\quad$ and we start from $U+V=2a$

Squaring we get $\quad U^2+2UV+V^2=4a^2$

$U^2+V^2=(x+c)^2+y^2+(x-c)^2+y^2=2(x^2+y^2+c^2)$

Now we'll divide by $2$ and put the roots on one side and square again:

$$U^2V^2=(2a^2-\tfrac 12(U^2+V^2))^2$$ $\begin{align}U^2V^2 &=((x+c)^2+y^2)((x-c)^2+y^2)\\ &=((x+c)^2+(x-c)^2)y^2+(x+c)^2(x-c)^2+y^4\\ &=(2x^2+2c^2)y^2+(x^2-c^2)^2+y^4\\ \require{cancel}&=\cancel{2x^2y^2}+\cancel{2c^2y^2}+\cancel{x^4}-2c^2x^2+\cancel{c^4}+\cancel{y^4} \end{align}$

It is equal to:

$(2a^2-x^2-y^2-c^2)^2 \require{cancel}=4a^4+\cancel{x^4}+\cancel{y^4}+\cancel{c^4}-4a^2x^2-4a^2y^2-4a^2c^2+\cancel{2x^2y^2}+2x^2c^2+\cancel{2y^2c^2}$

Gathering remaining terms:

$-2x^2c^2=4a^4-4a^2x^2-4a^2y^2-4a^2c^2+2x^2c^2\iff 4\Big(a^4-a^2x^2-a^2y^2-a^2c^2+x^2c^2\Big)=0$

$$(-a^2+c^2)x^2-a^2y^2+a^2(a^2-c^2)=0$$

Which is the desired expression.