JEE Main 2020MathematicsEllipseMediumMCQ

JEE Main 2020 — Ellipse Question with Solution

JEE Main 2020 (05 Sep Shift 1)

Question

If the point P on the curve, 4x2+5y2=20 is farthest from the point Q0,-4, then PQ2 is equal to

Choose an option

Show full solutionCorrect option: A
Correct answer
A36

Step-by-step explanation

Give ellipse

x25+y24=1       ...1

Let P5cosθ, 2sinθ

Ellipse (1)

Given Q0, -4

PQ2=5cosθ2+2sinθ22

⇒PQ2=5cos2θ+4sin2θ+16cos2θ+16

⇒PQ2=4+cos2θ+16 sinθ+16

=1-sin2θ+16 sinθ

⇒PQ2=17-sin2θ+16 sinθ

⇒PQ2=17-sin2θ-16 sinθ

⇒PQ2=21sinθ-82-64

⇒PQ2=85-(sinθ-8)2=85-49

Max PQ2=36       (∵ maximum value of sinθ=1)

 

 

 

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About this question

This is a previous-year question from JEE Main 2020, covering the Ellipse chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.