If $$P \equiv \left( {x,\,y} \right),\,{F_1} \equiv \left( {3,\,0} \right),\,{F_2} \equiv \left( { - 3,\,0} \right)$$        and $$16{x^2} + 25{y^2} = 400,$$     then $$P{F_1} + P{F_2}$$   equals :

Solution :
The ellipse can be written as, $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1$$
Here $${a^2} = 25,\,{b^2} = 16,$$
$$\eqalign{ & {\text{But }}{b^2} = {a^2}\left( {1 - {e^2}} \right) \cr & \Rightarrow \frac{{16}}{{25}} = 1 - {e^2} \cr & \Rightarrow {e^2} = 1 - \frac{{16}}{{25}} = \frac{9}{{25}} \cr & \Rightarrow e = \frac{3}{5} \cr} $$
Foci of the ellipse are $$\left( { \pm ae,\,0} \right) = \left( { \pm 3,\,0} \right),$$     i.e., $${F_1}$$ and $${F_2}$$
$$\therefore $$  We have $$P{F_1} + P{F_2} = 2a = 10$$     for every point $$P$$ on the ellipse.