JEE Main 2021MathematicsParabolaMediumNumerical

JEE Main 2021 — Parabola Question with Solution

JEE Main 2021 (20 Jul Shift 1)

Question

Let y=mx+c,m>0 be the focal chord of y2=-64x, which is tangent to (x+10)2+y2=4. Then, the value of 42( m+c) is equal to______

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Show full solutionCorrect answer: 34
Correct answer
34

Step-by-step explanation

We have,

y2=-64x

Coordinates of focus of parabola is : -16,0

y=mx+c is focal chord, hence

⇒c=16m  ...1

y=mx+c is tangent to (x+10)2+y2=4, therefore

y=mx+10±21+m2

⇒c=10 m±21+m2

⇒16 m=10 m±21+m2

⇒6m=21+m2,  m>0

⇒9 m2=1+m2

⇒m=122 & c=82

Then,

42m+c=421722=34

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About this question

This is a previous-year question from JEE Main 2021, covering the Parabola 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.