Question

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

• A. $8$
• B. $6$
• C. $10$
• D. $12$

Answer 1

The correct option is C $10$

The equation of the ellipse can be written as $\frac{x^2}{25}+\frac{y^2}{16}=1$

Here $a^2=25$, $b^2=16$

But $b^2=a^2(1-e^2)$

$\Rightarrow$ $16=25(1-e^2)$ $\Rightarrow$ $e=\frac{3}{5}$

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

By definition of the ellipse, since $P$ is any point on the ellipse

$P{F}_1+P{F}_2=2a=2\times 5=10$