JEE Main 2020MathematicsEllipseMediumMCQ

JEE Main 2020 — Ellipse Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

Let x2a2+y2b2=1a>b be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, ϕt=512+t-t2, then a2+b2 is equal to :

Choose an option

Show full solutionCorrect option: C
Correct answer
C126

Step-by-step explanation

LR=2b2a=10 ⇒   b2=5a

ϕt=512−t2-t+14−14=512+14−t−122

=23−t−122

max ϕt=23=e

b2=a21−e2

5a=a21−49

⇒  5=59a

⇒   a2=81,  b2=45

a2+b2=126.

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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.