JEE Main 2019MathematicsEllipseMediumMCQ

JEE Main 2019 — Ellipse Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

If the tangents on the ellipse 4x2+y2=8 at the points 1, 2 and (a, b) are perpendicular to each other, then a2 is equal to

Choose an option

Show full solutionCorrect option: A
Correct answer
A 217

Step-by-step explanation

4x2+y2=8
⇒8x+2ydydx=0
dydx=-8x2y=-4xy
∴ Slope at 1, 2, m1=-2
      Slope at a,b, m2=-4ab
∴ -2×-4ab=-1
       b=-8a
Again,
     4a2+b2=8
⇒4a2+64a2=8
⇒68a2=8
⇒a2=868

⇒a2=217

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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.