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.


Note that $y=0$ gives $x=\pm5$, so $-5 \le p \le 5$ for all points $P(p,q)$ on the ellipse. Therefore:
$$
|F_1P| = \frac{1}{5}\sqrt{(4p+25)^2} = \frac{1}{5}|4p+25| = \frac{1}{5}(25+4p) \\
|F_2P| = \frac{1}{5}\sqrt{(4p-25)^2} = \frac{1}{5}|4p-25| = \color{red}{\frac{1}{5}(25-4p)}
$$
