Let P be the point on the parabola, y 2 = 8 x which is at a minimum distance from the center C of the circle…

Let P be the point on the parabola, y2=8x which is at a minimum distance from the center C of the circle  x2+y+62=1. Then the equation of the circle, passing through C and having its center at P is
  1. x2+y2-x4+2y-24=0
  2. x2+y2-4x+9y+18=0
  3. x2+y2-4x+8y+12=0
  4. x2+y2-x+4y-12=0

Solution

y2=8x is the equation of the given parabola. If P is a point at a minimum distance from '0,-6', then it should be normal to the parabola at P.

Normal to parabola y2=8x is 

y=mx-2·2·m-2·m3

It passes through 0,-6

 m3+2m-3=0m=1 

Pam2,-2am=P(2,-4)

Equation of circle with centre P and passes through 'C' is

x-22+y+42=8

x2+y2-4x+8y+12=0

Asked in: JEE Main 2016 (03 Apr)

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