How to prove distance from foci on an ellipse is equal to twice the semi-major axis (for specific ellipse) Prove that for any point (x,y) on the conic, the sum of the distances to the two foci is always twice the semi-major axis.
I know that this can be proven in general for all ellipses but the practice question specifically asks for this to be proven for $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1. I feel like I'm really close but I've managed to math myself into a corner somehow.
Let the foci ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0) be denoted as F and F'. Let the point on the conic be denoted P(x,y). We are required to show PF + PF' = 2a. In this case, since a = 3, 2a = 6.
PF = $\sqrt{(x-\sqrt{5})^2 + y^2}$ and PF' = $\sqrt{(x+\sqrt{5})^2 + y^2}$
By rearranging the equation for the ellipse, we get y$^2$ = 4 - $\frac{4}{9}$x$^2$.
Substitute this into PF and PF' to get:
PF = $\sqrt{(x-\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 - 2\sqrt{5}x + 9}$ = $\sqrt{(x - \frac{9\sqrt{5}}{5})^2}$ = x - $\frac{9\sqrt{5}}{5}$
PF' = $\sqrt{(x+\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ = x + $\frac{9\sqrt{5}}{5}$
Therefore PF + PF' = 2x
And then I got stuck

An ellipse is a plane curve surrounding two foci, such that for all points on the curve, the sum of the two distances to the foci is a constant. One starts with $\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$ (your question) to arrive at $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $2a$ is any constant (which ends up being the length of the semi-major axis), $b^2=a^2-c^2$, and the foci are $(-c,0),(+c,0)$. Note that $a,b,c\in\mathbb R^+$.

Addendum
Finding the distance of either foci from a point $P(x,y)$ on the ellipse, $$PF=\sqrt{(x-c)^2+y^2}=\sqrt{(x-c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2-2cx+a^2}=\left\lvert\frac ca x-a\right\rvert=a-ex$$
$$PF^\prime=\sqrt{(x+c)^2+y^2}=\sqrt{(x+c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2+2cx+a^2}=\left\lvert\frac ca x+a\right\rvert=a+ex$$
since $x\in[-a,+a]$, where $e=\frac ca$.
