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

P is a point on $x^2=4y\Rightarrow p=\left(2t,t^2\right)$

           centre of the circle $x^2+y^2+6x+8=0$ $$

           is $c=\left(-3,0\right);pc=\sqrt{{\left(2t+3\right)}^2+t^4}$

             let $s=t^4+{\left(2t+3\right)}^2$

$\frac{ds}{dt}=4t^3+4\left(2t+3\right)$

$\frac{d^{2}s}{a{t}^2}=12t^2+8>0$

$\frac{ds}{dt}=0\Rightarrow 4t^3+4\left(2t+3\right)=0$

$t^3+2t+\text{}3=0\Rightarrow t=-1\therefore p=\left(-2,1\right)$

$x^2=4y;$$2x=4\frac{dy}{dx}\Rightarrow m=\frac{x}{2}=-1$

The equation of tangent at p is 

$y-1=-\left(x+2\right)$ 

$x+y+1=0$