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

For such a question, the realm of complex numbers helps a lot.
In the complex plane, an ellipse is described by :
$$|z-a| + |z-b|=c$$
where z represents each point on the ellipse, (a,b) is a tuple of complex numbers (which may or may not be real), c is an arbitrary constant $\geq |a-b|$
Also note, (a,b) are the focii of the ellipse as well and c is the major axis.
Coming on to the question, just describe the given equation in the form of complex numbers, the result it direct.

So, the given ellipse is  $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1 and
the focii ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0).
Describe this as
$$|z-\sqrt{5}| + |z-\sqrt{-5}|=6$$
This directly means that the sum of distance of any point $z$ on the ellipse is equal to the major axis.
