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 by definition the locus of the points such that the sum of their distances to the foci is constant.
From this property, the result is immediate:
$$PQ=FQ+QG.$$

