Prove that for a given point on an ellipse, the sum of the distances from each focal point is constant 
An ellipse has equation $\frac{x^2}{25}+\frac{y^2}{9} = 1$ and $P(p,q)$ is a point on the ellipse. Points $F_1$ and $F_2$ have coordinates $(-4,0)$ and $(4,0)$. Show that the sum of the distances $|PF_1|$ +$|PF_2|$ does not depend on the value of p.


First, to find  distance $|F_1P|$
$|F_1P| = \sqrt{(p+4)^2 + q^2}$
as $\frac{p^2}{25}+\frac{q^2}{9} = 1$
$\frac{9p^2}{25}+q^2 = 9$
$q^2 = 9-\frac{9p^2}{25}$
$|F_1P| = \sqrt{p^2+8p+16 + 9-\frac{9p^2}{25}}$
$= \sqrt{\frac{16}{25}p^2+8p-25}$
$= \sqrt{\frac{1}{25}(16p^2+200p-625)}$
$= \frac{1}{5}\sqrt{(4p+25)^2}$
$= \frac{1}{5}(4p+25)$
Finding the distance $|F_2P|$
$|F_2P| = \sqrt{(4-p)^2 + q^2}$
$= \sqrt{p^2-8p+16 + 9-\frac{9p^2}{25}}$
$= \sqrt{\frac{16}{25}p^2-8p-25}$
$= \sqrt{\frac{1}{25}(16p^2-200p-625)}$
$= \frac{1}{5}\sqrt{(4p-25)^2}$
$= \frac{1}{5}(4p-25)$
Therefore 
$|F_1P|+|F_2P|= \frac{1}{5}(4p-25)+\frac{1}{5}(4p+25) = \frac{8}{5}p$
This is exactly what we're trying to disprove. I must have made a mistake somewhere. Please can someone explain where I went wrong. My answer appears to be the negative of what it should be. I expect the correct answer will be 10 units but for some reason it does not seem to work.


The key mistake you made here is your simplification of square roots.
Actually, since a square root is positive always, you should get $\vert F_1P\vert = \frac{|4p+25|}{5} $ and $ \vert F_2P\vert = \frac{|4p-25|}{5}$.
However, note $ -5 \leq p \leq 5$, so $ -20 \leq 4p \leq 20$ and $\frac{|4p+25|}{5} = \frac{4p+25}{5}, \frac{|4p-25|}{5} = \frac{-4p+25}{5}$ Summing the two distances, $p$ cancels out.
