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=0is minimum, then the equation of the tangent to the parabola at P, is

Let $P (2 t, t^{2})$

Given, $x^{2} + y^{2} + 6 x + 8 = 0$ is equation of the circle.

So, coordinates of centre is $(- 3, 0)$ .

Now, distance of point P from $(- 3, 0)$

$= \sqrt{(2 t + 3)^{2} + {(t^{2} - 0)}^{2}}$

Let $f (t) = (2 t + 3)^{2} + {(t^{2} - 0)}^{2} = 4 t^{2} + 9 + 12 t + t^{4}$ …(i)

Differentiating equation (i) w.r.t. ‘t’, we have

$f^{'} (t) = 8 t + 12 + 4 t^{3} = 4 (t^{3} + 2 t + 3) = 4 (t + 1) (t^{2} - t + 3)$

Now, $f^{'} (t) = 0 \Rightarrow t = - 1$

So, point P becomes (-2,1)

$∴$ Equation of tangent to the parabola is given by

$x x_{1} = 2 (y + y_{1}) \Rightarrow x (- 2) = 2 (y + 1) \Rightarrow x + y + 1 = 0$

Let P be a point on the parabola, $x^{2} = 4 y$ . If the distance of P from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at P, is