JEE Main 2019MathematicsEllipseMediumMCQ

JEE Main 2019 — Ellipse Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

Let O0,0 and A0,1 be two fixed points. Then, the locus of a point P such that the perimeter of ΔAOP is 4 is

Choose an option

Show full solutionCorrect option: D
Correct answer
D9x2+8y2-8y=16

Step-by-step explanation


 Given, OA+PO+PA=4 

⇒PO+PA=3 

⇒Locus of P is ellipse with foci at O & A and major

axis 2b=3

Distance between foci =2be=1⇒e=13

⇒Minor axis 2a=2b1-e2=3.223=22

⇒ Equation of Locus of P is :

x28+y-1229=14

By simplifying the above equation, we get

9x2+8y2-8y=16

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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.