If the co-ordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic $9 x^{2}+16 y^{2}=144$, then $P A+P B$ is equal to:

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Ellipse

If the co-ordinates of two points

A

and

B

are

(7, 0)

and

(- 7, 0)

respectively and

P

is any point on the conic

9 x^{2} + 16 y^{2} = 144

, then

P A + P B

is equal to:

Answer: (A)

Write the conic as

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

, an ellipse with

a = 4

and

b = 3

. Its eccentricity is

e = 1 - \frac{9}{16} = \frac{7}{4}

, so the foci lie at

(\pm a e, 0) = (\pm 7, 0)

, which are exactly the points

A

and

B

. By the defining property of the ellipse the sum of the focal distances is constant and equal to

2 a

, so

P A + P B = 8

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option