JEE Main 2022MathematicsHyperbolaEasyMCQ

JEE Main 2022 — Hyperbola Question with Solution

JEE Main 2022 (25 Jul Shift 2)

Question

Let the foci of the ellipse x216+y27=1 and the hyperbola x2144-y2α=125 coincide. Then the length of the latus rectum of the hyperbola is:

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Show full solutionCorrect option: D
Correct answer
D2710

Step-by-step explanation

Given equation of ellipse x216+y27=1

Now finding eccentricity =1-716=34

So, foci ≡±ae,0≡±3,0

Now, hyperbola: x214425-y2α25=1

Eccentricity will be =1+α144=112144+α

Foci ≡±ae,0≡±125·112144+α,0

Given foci coincide then 3=15144+α⇒α=81

Hence, hyperbola is x21252-y2952=1

Length of latus rectum =2b2a=2·8125125=2710

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About this question

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