JEE Main 2020MathematicsHyperbolaMediumMCQ

JEE Main 2020 — Hyperbola Question with Solution

JEE Main 2020 (03 Sep Shift 2)

Question

Let e1 and e2 be the eccentricities of the ellipse x225+y2b2=1 b<5 and the hyperbola x216-y2b2=1 respectively satisfying e1e2=1. If α and β are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair α, β is equal to:

Choose an option

Show full solutionCorrect option: A
Correct answer
A8, 10

Step-by-step explanation

e1=1-b225 and e2=1+b216

Given e1e2=1

⇒e1e22=1⇒1-b2251+b216=1

⇒b216-b225-b425×16=0

⇒916×25b2-b425×16=0⇒b2=9

Thus, e1=1-925=45 and e2=1+916=54

Therefore, α=25e1=8 and β=24e2=10

∴α, β=8, 10

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About this question

This is a previous-year question from JEE Main 2020, 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.