JEE Main 2020MathematicsEllipseMediumMCQ

JEE Main 2020 — Ellipse Question with Solution

JEE Main 2020 (07 Jan Shift 2)

Question

If 3x+4y=122 is a tangent o the ellipse x2a2+y29=1 for some a∈R, then the distance between the foci of the ellipse is

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Show full solutionCorrect option: A
Correct answer
A27

Step-by-step explanation

3x+4y=122
⇒4y=-3x+122
⇒y=-34x+32
Condition of tangency c2=a2m2+b2
18=a2∙916+9
a2∙916=9
a2=16
a=4
e=1-b2a2=1-916=74
∴ae=74∙4=7
∴ Focus are ±7,0
∴ Distance between foci =27

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About this question

This is a previous-year question from JEE Main 2020, 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.