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

$\text{The ellipse can be written as}$

$\begin{array}{l}\frac{x^2}{25}+\frac{y^2}{16}=1\\ \text{ Here, }{a}^2=25,b^2=16.\text{ Now, }\\ b^2=a^2\left(1-e^2\right)\\ \text{ or }\frac{16}{25}=1-e^2\\ \text{ or }{e}^2=1-\frac{16}{25}=\frac{9}{25}\\ \text{ or }e=\frac{3}{5}\end{array}$

$\text{ The foci of the ellipse are }(\pm ae,0)\equiv (\pm 3,0)\text{ , i.e., }{F}_1\text{ and }{F}_2\text{ are the foci of the ellipse. }$

$\text{ Therefore, we have }P{F}_1+P{F}_2=2a=10\text{ for every point }P\text{ on the ellipse.}$