Ellipse Ques 11
- If $P=(x, y), F _1=(3,0), F _2=(-3,0)$ and $16 x^{2}+25 y^{2}=400$, then $P F _1+P F _2$ equals $(1998,2 M)$
(a) 8
(b) 6
(c) 10
(d) 12
Show Answer
Answer:
Correct Answer: 11.(c)
Solution:
- Given, $16 x^{2}+25 y^{2}=400$ [given]
$\Rightarrow \quad \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$
Now, $\quad P F _1+P F _2=$ Major axis $=2 a$ [where, $a=5$ ]
$ =2 \times 5=10 $
BETA
