JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Ellipse

Let the maximum area of the triangle that can be inscribed in the ellipse

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

a > 2

, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be

63

. Then the eccentricity of the ellipse is:

Answer: (A)

Solution

Let one vertex be at

(a, 0)

and another at

(a cos θ, 2 sin θ)

on the ellipse.

The area of the triangle is:

A = \frac{1}{2} \cdot a (1 - cos θ) \cdot 4 sin θ = 2 a (1 - cos θ) sin θ

Differentiating:

\frac{d A}{d θ} = 2 a (sin^{2} θ + cos θ - cos^{2} θ) = 2 a (1 + cos θ - 2 cos^{2} θ)

Setting to zero:

2 cos^{2} θ - cos θ - 1 = 0

, giving

cos θ = - \frac{1}{2}

(or

θ = \frac{2 π}{3}

)

Maximum area:

A_{m a x} = 2 a \cdot \frac{3}{2} \cdot \frac{3}{2} = \frac{3 3 a}{2} = 63

a = 4

.

Eccentricity:

e = \frac{a ^{2} - b ^{2}}{a ^{2}} = \frac{16 - 4}{16} = \frac{3}{2}

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option