Question

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

• A. $8$
• B. $6$
• C. $10$
• D. $12$

Answer 1

Answer: (C)

$10$

Given equation is $16x^2+25y^2=400$

The ellipse can be written as, $\frac{x^2}{25}+\frac{y^2}{16}=1$
Here $a^2=25,b^2=16$, but $b^2=a^2(1-e^2)$
$\Rightarrow \frac{16}{25}=1-e^2$
$\Rightarrow e^2=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow e=\pm \frac{3}{5}$
Foci of the ellipse are $(+ae,0)$ $=$ $(\pm 3,0)$, i.e., $F_1$ and $F_2$.
We have $P{F}_1+P{F}_2=2a=10$ for every point $P$ on the ellipse.