JEE Main 2022MathematicsEllipseEasyMCQ

JEE Main 2022 — Ellipse Question with Solution

JEE Main 2022 (26 Jun Shift 2)

Question

If m is the slope of a common tangent to the curves x216+y29=1 and x2+y2=12, then 12m2 is equal to

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Show full solutionCorrect option: B
Correct answer
B9

Step-by-step explanation

Given,

x216+y29=1

we know that equation of tangent to the ellipse is

y=mx±a2 m2+b2

y=mx±16m2+9     ...1

x2+y2=12

Also equation of tangent to the circle is

y=mx±121+m2     ...2

Now, for common tangent equate eq.1 and 2

⇒16m2+9=121+m2

16m2-12m2=3

4m2=3

12m2=9

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About this question

This is a previous-year question from JEE Main 2022, 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.