JEE Main 2024MathematicsHyperbolaEasyMCQ

JEE Main 2024 — Hyperbola Question with Solution

JEE Main 2024 (01 Feb Shift 1)

Question

Let x2a2+y2b2=1,a>b be an ellipse, whose eccentricity is 12 and the length of the latus rectum is 14. Then the square of the eccentricity of x2a2−y2b2=1 is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C32

Step-by-step explanation

Given: Length of latus rectum of ellipse is 14.

⇒2b2a=14   ...i

Also, eccentricity is given as 12

⇒12=1−b2a2

⇒12=1−b2a2

⇒12=1−a142a2

⇒-12=−142a

⇒a=14

⇒a2=14, b2=7

Now, eH=1+b2a2

⇒eH=1+12

⇒eH=32

⇒eH2=32

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About this question

This is a previous-year question from JEE Main 2024, 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.