JEE Main 2022MathematicsHyperbolaEasyNumerical

JEE Main 2022 — Hyperbola Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

Let H:x2a2-y2 b2=1,a>0, b>0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 422+14. If the eccentricity H is 112, then value of a2+b2 is equal to ______.

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Show full solutionCorrect answer: 88
Correct answer
88

Step-by-step explanation

Given x2a2-y2b2=1, 2a+2b=422+14 and e=112

We know  e2=1+b2a2⇒114=1+b2a2⇒b2=74a2

∴x2a2-y272a2=1 

Now 2a+2·7a2=422+14

a2+7=422+7

a=42⇒a2=32

b2=74×16×2=56

Hence a2+b2=88

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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.