If $P(x,y),F_1=(3,0),F_2=(-3,0)$ and $16x^2+25y^2=400$ then $P{F}_1+P{F}_2$ equals:

We have $16x^2+25y^2=400\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$ or $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=25$ and $b^2=16$ This equation represents an ellipse with eccentricity given by $e^2=1-\frac{b^2}{a^2}=1-\frac{16}{25}=\frac{9}{25}\Rightarrow e=3/5$

So, the coordinates of the foci are $(\pm ae,0)$ i.e. (3,0) and (-3,0), Thus $F_1$ and $F_2$ are the foci of the ellipse.

Since, the sum of the focal distance of a point on an ellipse is equal to its major axis,

$\therefore P{F}_1+P{F}_2=2a=10$