JEE Main 2018MathematicsEllipseMediumMCQ

JEE Main 2018 — Ellipse Question with Solution

JEE Main 2018 (15 Apr)

Question

If β is one of the angles between the normals to the ellipse x2+3y2=9 at the points 3cosθ,3sinθ and -3sinθ,3cosθ; θ∈0,π2; then 2cotβsin2θ is equal to :

Choose an option

Show full solutionCorrect option: C
Correct answer
C23

Step-by-step explanation

Given equation of ellipse is x29+y23=1

Normal at 3cosθ,3sinθ is

3secθ.x-3cosecθy=6

⇒Slope m1=3secθ3cosecθ=3tanθ

Normal at −3sinθ,3cosθ

-3cosecθ.x-3secθ.y=6

⇒Slope m2=-3cosecθ3secθ=-3cotθ

Angle between normal is β.

⇒tanβ=m1-m21+m1m2=3tanθ+3cotθ1-3=32sinθcosθ

⇒tanβ=3sin2θ⇒2cotβsin2θ=23

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About this question

This is a previous-year question from JEE Main 2018, 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.