JEE Main 2019MathematicsHyperbolaMediumMCQ

JEE Main 2019 — Hyperbola Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

Let 0<θ<π2. If the eccentricity of the hyperbola x2cos2⁡θ-y2sin2⁡θ=1 is greater than 2, then the length of its latus rectum lies in the interval:

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Show full solutionCorrect option: A
Correct answer
A3, ∞

Step-by-step explanation

For given hyperbola the eccentricity is given as

e2=1+sin2θcos2θ=1+tan2θ=sec2θ

⇒e=secθ

∴ Length of latus rectum l=2sin2θcosθ=2tan2θsecθ

⇒l=2(e2-1)e=2e-1e
On differentiating w.r.t. e, we get

dlde=21+1e2>0

∴ l is an increasing function

∴lmin=22-12=3

∴  Range of latus rectum is 3,∞.

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About this question

This is a previous-year question from JEE Main 2019, 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.