JEE Main 2017MathematicsEllipseEasyMCQ

JEE Main 2017 — Ellipse Question with Solution

JEE Main 2017 (02 Apr)

Question

The eccentricity of an ellipse whose centre is at the origin is 12 . If one of its directrices is x=-4 , then the equation of the normal to it at 1,32 is:

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Show full solutionCorrect option: B
Correct answer
B4x-2y=1

Step-by-step explanation

Given, e=12 and ae=4   

⇒a=2, 

Also, b2=a21-e2

⇒b2=3

 Equation of ellipse is  x24+y23=1   
We know that, equation of normal to the ellipse x2a2+y2b2=1 at x1, y1 is a2xx1-b2yy1=a2-b2
So, equation of normal at 1, 32 is  4x1-3y3/2=4-3 
⇒4x-2y=1

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About this question

This is a previous-year question from JEE Main 2017, 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.