JEE Main 2022MathematicsEllipseEasyNumerical

JEE Main 2022 — Ellipse Question with Solution

JEE Main 2022 (27 Jul Shift 1)

Question

If the length of the latus rectum of the ellipse x2+4y2+2x+8y-λ=0 is 4, and l is the length of its major axis, then λ+l is equal to _____.

Enter your answer

Show full solutionCorrect answer: 75
Correct answer
75

Step-by-step explanation

Given, 

x2+4y2+2x+8y-λ=0 is a ellipse

Now on rearranging the terms we get,

⇒x+12λ+5+y+12λ+54=1

We know that length of latus rectum is 2b2a=4 

So, 2λ+54λ+5=4

⇒2λ+54=4λ+5

⇒λ+5=8λ+5

⇒λ+52=64λ+5

⇒λ2+25+10λ=64λ+5

⇒λ2-54λ-295=0

⇒ λ=59 or -5

But λ≠-5

So, l=2a=2λ+5=264=16

⇒ λ+l=59+16=75

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Ellipse chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, 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.