Verify that an equation is an ellipse I want to show that the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is an ellipse with foci $(\pm\sqrt{a^2-b^2},0)$.
I started with summing the two distances and try to prove a constant but realise it is very tedious.
$\sqrt{(x-\sqrt{a^2-b^2})^2+y^2}+\sqrt{(x+\sqrt{a^2-b^2})^2+y^2}=...$
I tried considering proving the square of this quantity is a constant but I can't get anywhere either.
But by the geometrical definition of ellipse the quantity above must be a constant right?

It's important to realize that we can plug in $y^2$ from the equation of the ellipse. Solving for $y^2$ in the ellipse, we get $$y^2 = b^2-\frac{b^2x^2}{a^2}$$ Then we plug this in to get $$\sqrt{(x-\sqrt{a^2-b^2})^2+b^2-\frac{b^2x^2}{a^2}}+\sqrt{(x+\sqrt{a^2-b^2})^2+b^2-\frac{b^2x^2}{a^2}}=$$
$$\sqrt{x^2-2x\sqrt{a^2-b^2}+a^2-\frac{b^2x^2}{a^2}}+\sqrt{x^2+2x\sqrt{a^2-b^2}+a^2-\frac{b^2x^2}{a^2}}$$
Simplifying, we get $$\left|x\sqrt{1-\frac{b^2}{a^2}}-a\right|+\left|x\sqrt{1-\frac{b^2}{a^2}}+a\right|$$ Since we are only worried about the interval of $[-a, a]$, we can simplify this to $$a-x\sqrt{1-\frac{b^2}{a^2}}+x\sqrt{1-\frac{b^2}{a^2}}+a = 2a$$ This is a constant value, and thus, the equation is an ellipse.
