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JEE Advanced 2020 · Paper 1, Mathematics question 4

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=4λxy^2 = 4\lambda x, 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 is

Choose one option
Answer and worked solution

Official answer: A

The point. Take P=(λ,2λ)P=(\lambda,2\lambda), an end of the latus rectum.

Slopes at PP.

  • Parabola: dydx=2λy=1\dfrac{dy}{dx}=\dfrac{2\lambda}y=1.
  • Ellipse: dydx=−b2xa2y=−b22a2\dfrac{dy}{dx}=-\dfrac{b^2x}{a^2y}=-\dfrac{b^2}{2a^2}.

Perpendicular tangents.
1⋅(−b22a2)=−1  ⟹  b2=2a2.1\cdot\left(-\frac{b^2}{2a^2}\right)=-1\implies b^2=2a^2.
So b>ab>a and the major axis is along yy:
e2=1−a2b2=12  ⟹  e=12.e^2=1-\frac{a^2}{b^2}=\frac12\implies e=\frac1{\sqrt2}.
(A)

Topics:EllipseParabolaTangents and normals to conics
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