JEE Main 2019MathematicsEllipseHardMCQ

JEE Main 2019 — Ellipse Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

If the normal to the ellipse 3x2+4y2=12 at a point P on it is parallel to the line, 2x+y=4 and the tangent to the ellipse at P passes through Q(4,4) then PQ is equal to:

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Show full solutionCorrect option: B
Correct answer
B55 2

Step-by-step explanation

Given the equation of an ellipse is x24+y23=1,where a=2, b=3.
As we know equation of normal to the ellipse x2a2+y2b2=1 can be taken as
axsec⁡θ-bycosec⁡θ=a2-b2
⇒2xsec⁡θ-3ycosec⁡θ=1 for the given ellipse.
Now, this normal is parallel to 2x+y=5.
⇒-2sec⁡θ3cosec⁡θ=-2⇒tan⁡θ=-3⇒θ=2π3,-π3
Hence point Pacos⁡θ, bsin⁡θ=-1,32 or 1, -32
Tangent at -1,32 is x.-14+y.323=1
⇒-x+2y=4 which also passes through Q4, 4
⇒PQ=4±12+4∓322=552.

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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.