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

We are asked to find that point $$P(x,y)$$ on the parabola $$x^{2}=4y$$ which is nearest to the centre of the circle $$x^{2}+y^{2}+6x+8=0$$, and then to write the equation of the tangent to the parabola at that particular point.

First we rewrite the equation of the circle in its standard (centre-radius) form. We have

$$x^{2}+y^{2}+6x+8=0.$$

To complete the square in $$x$$ we add and subtract $$9$$ (because $$(x+3)^{2}=x^{2}+6x+9$$). So

$$x^{2}+6x+9+y^{2}= -8+9,$$

which becomes

$$(x+3)^{2}+y^{2}=1.$$

Hence the centre of the circle is $$C(-3,0)$$ and its radius is $$r=1$$ (though we shall actually need only the centre).

Now we choose a general point $$P$$ on the parabola. For the parabola $$x^{2}=4y$$ it is convenient to take

$$x=t,\qquad y=\dfrac{t^{2}}{4},$$

where $$t$$ is a real parameter. Thus

$$P(t,\;t^{2}/4).$$

The squared distance between $$P$$ and the fixed point $$C(-3,0)$$ is

$$D^{2}=(x+3)^{2}+(y-0)^{2}.$$

Substituting $$x=t$$ and $$y=\dfrac{t^{2}}{4}$$ gives

$$D^{2}=(t+3)^{2}+\left(\dfrac{t^{2}}{4}\right)^{2}.$$

Let us expand and tidy this expression because minimising $$D^{2}$$ is the same as minimising $$D$$ itself.

First expand $$(t+3)^{2}$$:

$$(t+3)^{2}=t^{2}+6t+9.$$

Next, square the second term:

$$\left(\dfrac{t^{2}}{4}\right)^{2}=\dfrac{t^{4}}{16}.$$

Therefore

$$D^{2}=t^{2}+6t+9+\dfrac{t^{4}}{16}.$$

Define

$$f(t)=\dfrac{t^{4}}{16}+t^{2}+6t+9.$$

We differentiate $$f(t)$$ with respect to $$t$$ and set the derivative to zero to locate the critical point. Using the power rule $$\dfrac{d}{dt}(t^{n})=nt^{n-1}$$, we obtain

$$\frac{d}{dt}\!\left(\dfrac{t^{4}}{16}\right)=\dfrac{4t^{3}}{16}=\dfrac{t^{3}}{4},$$ $$\frac{d}{dt}(t^{2})=2t,$$ $$\frac{d}{dt}(6t)=6,$$ and of course the derivative of the constant $$9$$ is $$0$$.

Hence

$$f'(t)=\dfrac{t^{3}}{4}+2t+6.$$

Setting $$f'(t)=0$$ gives

$$\dfrac{t^{3}}{4}+2t+6=0.$$

Multiply through by $$4$$ (to clear the denominator) and we have

$$t^{3}+8t+24=0.$$

To solve this cubic we look for simple integral roots. Try $$t=-2$$:

$$(-2)^{3}+8(-2)+24=-8-16+24=0.$$

Because $$t=-2$$ satisfies the equation, $$(t+2)$$ is a factor. Carrying out polynomial division (or synthetic division) gives

$$t^{3}+8t+24=(t+2)\bigl(t^{2}-2t+12\bigr).$$

The quadratic factor $$t^{2}-2t+12$$ has discriminant $$\Delta=(-2)^{2}-4(1)(12)=4-48=-44<0,$$ so it yields no real roots. Therefore the only real critical point is

$$t=-2.$$

This critical point indeed minimises $$f(t)$$ (and hence $$D^{2}$$), so the required point $$P$$ is

$$x=-2,\qquad y=\dfrac{(-2)^{2}}{4}=\dfrac{4}{4}=1.$$

Thus

$$P(-2,1).$$

Next we need the tangent to the parabola $$x^{2}=4y$$ at this point. First we compute the slope using calculus. Differentiating the parabola implicitly with respect to $$x$$ gives

$$\frac{d}{dx}(x^{2})=\frac{d}{dx}(4y).$$

Because $$\frac{d}{dx}(x^{2})=2x$$ and $$\frac{d}{dx}(4y)=4\frac{dy}{dx},$$ we obtain

$$2x=4\frac{dy}{dx}.$$

Hence

$$\frac{dy}{dx}=\frac{x}{2}.$$

At the point $$(-2,1)$$ the slope is

$$m=\frac{-2}{2}=-1.$$

Using the point-slope form of a straight line, $$y-y_{1}=m(x-x_{1}),$$ with $$(x_{1},y_{1})=(-2,1)$$ and $$m=-1,$$ we write

$$y-1=-1\,(x+2).$$

Expand the right‐hand side:

$$y-1=-x-2.$$

Now bring all terms to one side:

$$x+y+1=0.$$

Thus the tangent to the parabola at the nearest point is

$$x+y+1=0.$$

Comparing with the options, this corresponds to Option A.

Hence, the correct answer is Option A.