Question

If $P = (x, y),F_1 = (3,0), F_2 = ( 3, 0)$ and $ 16x^2 + 25y^2 = 400,$ then $PF_1 + PF_2$ equals

• A. 8
• B. 6
• C. 10
• D. 12

Answer 1

Answer: (C)

Given, $16 x^2 + 25y^2 = 400$ [given]
$\implies \frac{x^2}{25}+ \frac{y^2}{16} = 1$
Here, $ a^2 = 25, b^2 = 16$
But $ b^2 = a^2 (1- e^2)$
$\implies 16= 25 (1-e^2) \implies \frac {16}{25}= 1- e^2$
$\implies e^2 = 1 - \frac {16}{25} = \frac {9}{25}\implies e = \frac {3}{5}$
Now, foci of the ellipse are $(\pm ae, 0) = (\pm 3 ,0).$
We have, $ 3 = a \frac {3}{5} \implies a = 5 $ Now, $PF_1 + PF_2 =$ Major axis $= 2a $
$ = 2 \times 5 = 10$