JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Ellipse

Let P be a point on the ellipse

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

. Let the line passing through P and parallel to y -axis meet the circle

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

at point

Q

such that

P

and

Q

are on the same side of the x -axis. Then, the eccentricity of the locus of the point R on PQ such that

PR : RQ = 4 : 3

as P moves on the ellipse, is :

Answer: (D)

Solution

Parametrize the ellipse as

P = (3 cos θ, 2 sin θ)

. The vertical line through

P

meets the circle

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

(same side of the x-axis) at

Q = (3 cos θ, 3 sin θ)

R

divides

P Q

with

P R : R Q = 4 : 3

, so

R = P + \frac{4}{7} (Q - P) .

Since

P

and

Q

share the same x-coordinate, the x-coordinate of

R

is simply

h = 3 cos θ

.

For the y-coordinate:

k = 2 sin θ + \frac{4}{7} (3 sin θ - 2 sin θ) = 2 sin θ + \frac{4}{7} sin θ = \frac{18 sin θ}{7} .

cos θ = \frac{h}{3}

and

sin θ = \frac{7 k}{18}

. Using

sin^{2} θ + cos^{2} θ = 1

\frac{h ^{2}}{9} + \frac{49 k ^{2}}{324} = 1.

This is an ellipse with

a^{2} = 9

and

b^{2} = \frac{324}{49}

(and

b^{2} < a^{2}

, so the major axis is along the x-axis).

e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{324/49}{9} = 1 - \frac{36}{49} = \frac{13}{49} ⟹ e = \frac{13}{7} .

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option