JEE Main 2022MathematicsHyperbolaMediumNumerical

JEE Main 2022 — Hyperbola Question with Solution

JEE Main 2022 (24 Jun Shift 2)

Question

Let the hyperbola H:x2a2-y2=1 and the ellipse E:3x2+4y2=12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of 12eH2+eE2 is equal to _____.

Enter your answer

Show full solutionCorrect answer: 42
Correct answer
42

Step-by-step explanation

Given

E≡3x2+4y2=12  ⇒ x24+y23=1

Now,  eE=1-b2a2=1-34=12

Length of L.R. =2b2a=2×32=3

Now H≡  x2a2-y21=1

 Length of L.R.=2b2a=2×1a

Given length of L.R. of E and H are equal

So 2a=3  ⇒  a=23

Now eH=b2a2+1=1232+1  ⇒  eH=94+1=134

So 12eH2+eE2=12134+14=12144=14×3=42

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Hyperbola 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 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.