Question

Let p be a point on the parabola, x2 = 4y. If the distance of p from the centre of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at p, is :

Answer 1

Given: P be a point on the parabola, x^2 = 4y,  centre of the circle, x^2 + y^2 + 6x + 8 = 0 is minimum.

To find: The equation of the tangent to the parabola at P

Solution:

• Let the point P on parabola be (2t , t^2) and the centre be (-g, -f)

• Then the centre will be: (-3, 0)

• Now slope for line from centre to P will be:

             y2 - y1 / x2 - x1 = t^2 - 0 / 2t + 3       ...................(i)

• So the slope of tangent to the parabola will be :

             dy/dx = x / 2 = t

             Slope of normal is: -1/t                   .......................(ii)

• Now comparing (i) and (ii) we get:

              t^2 - 0 / 2t + 3  =   t

              t^3 + 2t + 3 = 0

              (t+1)(t^2 - t + 1) = 0

• So the real root will be: -1

• So the coordinates of P will be: (2t , t^2)  =  (-2 ,1 )

• Now slope of tangent to parabola at P will be:

              dy/dx = x/2 = t = -1

• So the equation of tangent will be:

              y-1 = -1(x+2)

Answer:

       So the equation of the tangent to the parabola at P is y + x - 3 = 0