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Mathematics · Ellipse

JEE Advanced 2020 — Paper 1 — Question 32

Let a,b\mathrm{a}, \mathrm{b} and λ\lambda be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y2=\mathrm{y}^{2}= 4λx4 \lambda x, and suppose the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{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

  1. Option A:

    12\frac{1}{\sqrt{2}}

    Correct
  2. Option B:

    12\frac{1}{2}

  3. Option C:

    13\frac{1}{3}

  4. Option D:

    25\frac{2}{5}

Answer: A

Step-by-step solution

Given a,b,λ∈R+\mathrm{a}, \mathrm{b}, \lambda \in \mathrm{R}^{+}

As P lies on ellipse λ2a2+yλ2 b2=1\frac{\lambda^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y} \lambda^{2}}{\mathrm{~b}^{2}}=1

So equation of tangent at PP to ellipse ss, m=−λa2×b22λm=\frac{-\lambda}{a^{2}} \times \frac{b^{2}}{2 \lambda}

(Slope of tangent for parabola at P ) (Slope of tangent at P for ellipse) =−1=-1

⇒−b22a2=−1, b2=2a2\Rightarrow-\frac{\mathrm{b}^{2}}{2 \mathrm{a}^{2}}=-1, \mathrm{~b}^{2}=2 \mathrm{a}^{2}

e=1−12=12e=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Mathematics
Chapter
Ellipse
Topic
Tangents & Normals to ellipse, chord of conatct
Let a , b and λ be positive real numbers. Suppose P is an end point… | JEE Advanced 2020 PYQ with Solution · DhiX AI