JEE Main 2019MathematicsParabolaMediumMCQ

JEE Main 2019 — Parabola Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

Equation of a common tangent to the circle, x2+y2-6x=0 and the parabola, y2=4x is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B3y=x+3

Step-by-step explanation

Let tangent to parabola y2=4x is

y=mx+1m...i

If equation (i) is tangent to given circle whose centre is (3, 0) and radius is 3 then length of perpendicular from centre of circle to equation (i) is equal to radius of circle.

Hence, 3m+1mm2+1=3

⇒3m2+1=3m1+m2

⇒9m4+6m2+1=9m2+9m4

⇒m=±13

∴ common tangents are y=x3+3 or y=-x3-3

⇒3y=x+3 or 3y=-x-3 

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About this question

This is a previous-year question from JEE Main 2019, 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.