JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Ellipse

Let a, b and

λ

be positive real numbers. Suppose P is an end point of the latus rectum of the parabola

y^{2} = 4 λ x

, and suppose the ellipse

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

passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is

(A) $\frac{1}{2}$ ✓ Correct
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D) $\frac{2}{5}$

Answer: (A)

Solution

For the parabola

y^{2} = 4 λ x

, the end point of the latus rectum is

P (λ, 2 λ)

.

The slope of the tangent to the parabola at P is:

\frac{d y}{d x} = \frac{4 λ}{2 y} = \frac{4 λ}{4 λ} = 1

.

For the ellipse

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

, differentiating:

\frac{2 x}{a ^{2}} + \frac{2 y y ^{'}}{b ^{2}} = 0

.

Since the tangents are perpendicular, the slope at P on the ellipse is

y^{'} = - 1

.

Substituting

P (λ, 2 λ)

\frac{2 λ}{a ^{2}} - \frac{4 λ}{b ^{2}} = 0

, which gives

\frac{a ^{2}}{b ^{2}} = \frac{1}{2}

.

Therefore,

e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{1}{2} = \frac{1}{2}

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Single correct option

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Ellipse

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