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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2020 · Paper 1 · Mathematics · Application of Derivatives

Let a, b and λ be positive real numbers. Suppose P is an end point...

Let a,ba, b and λ\lambda be positive real numbers. Suppose PP is an end point of the latus rectum of the parabola y2=4\lambdax{y}^{2} = 4\lambdax, and suppose the ellipse x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1 passes through the point PP. If the tangents to the parabola and the ellipse at the point PP are perpendicular to each other, then the eccentricity of the ellipse isLet a, b and λ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y ^ 2 = 4 , and

Options

  1. A12\dfrac{1}{\sqrt{2}}
  2. B12\dfrac{1}{2}
  3. C13\dfrac{1}{3}
  4. D25\dfrac{2}{5}

Correct answer

A. 12\dfrac{1}{\sqrt{2}}

Step-by-step solution

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 , and

y2=4\lambdax⇒(dydx)P=1=m1{y}^{2} = 4\lambdax ⇒ {\left(\dfrac{dy}{dx}\right)}_{P} = 1 = {m}_{1}

Also, x2a2+y2b2=1⇒(dydx)P=−b22a2=m2\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1 ⇒ {\left(\dfrac{dy}{dx}\right)}_{P} = \dfrac{- {b}^{2}}{2{a}^{2}} = {m}_{2}

Given both tangents are perpendicular.

⇒m1⋅m2=−1⇒b2=2a2⇒ {m}_{1} ⋅ {m}_{2} = - 1 ⇒ {b}^{2} = 2{a}^{2}

And a2=b2(1−e2){a}^{2} = {b}^{2}\left(1 - {e}^{2}\right)

⇒1=2(1−e2)⇒ 1 = 2\left(1 - {e}^{2}\right)

⇒e=12⇒ e = \dfrac{1}{\sqrt{2}}

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