Let P be a point on the parabola, x2 = 4y.

Question

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

Accepted Answer

Let P(2t, t2) be any point on the parabola.

Center of the given circle C = (- g, -f) = (-3, 0)

For PC to be minimum, it must be the normal  to the parabola at P.

Slope of line PC = $${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \over {2t + 3}}$$

Also, slope of tangent to parabola at P = $${{dy} \over {dx}}$$ = $${x \over 2}$$ = t

$$ \therefore $$ Slope of normal = $${{ - 1} \over t}$$

$$ \therefore $$ $${{{t^2} - 0} \over {2t + 3}}$$ = $${{ - 1} \over t}$$

$$ \Rightarrow $$ t3 + 2t + 3 = 0

$$ \Rightarrow $$ (t+1) (t2 -  t + 3) = 0

$$\therefore\,\,\,$$ Real roots of above equation is

t = - 1

Coordinate of P = (2t, t2) = (-2, 1)

Slope of tangent to parabola at P = t = - 1

Therefore, equation of tangent is :

(y - 1) = (- 1) (x + 2)

$$ \Rightarrow $$  x + y + 1 = 0

Solutions · 1