JEE Main 2019MathematicsEllipseEasyMCQ

JEE Main 2019 — Ellipse Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

If the line x-2y=12 is a tangent to the ellipse x2a2+y2b2=1 at the point 3,-92, then the length of the latus rectum of the ellipse is

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Show full solutionCorrect option: C
Correct answer
C9 units

Step-by-step explanation

∵ 3,-92 lies on x2a2+y2b2=1⇒9a2+814b2=1 ……1

Equation of the tangent at 3,-92 is 3xa2+-92yb2=1

& given equation of the tangent is:

x-2y=12 ⇒x12+ -y6=1

On comparing these equations:

a23=12⇒a2=36⇒a=6

2b29=6⇒b2=27⇒b=33

Therefore, the length of latus rectum 

=2b2a=2×276=9 

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About this question

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