If two tangents drawn from a point $$P$$ to the parabola $$y^2 = 16(x-3)$$ are at right angles, then the locus of point $$P$$ is:

We consider the parabola $$y^{2}=16\,(x-3)$$.

First we rewrite it in the standard form $$y^{2}=4a\,(x-3)$$ so that we can recognise the parameter $$a$$. Comparing with $$y^{2}=4aX$$ we obtain $$4a=16 \Rightarrow a=4$$. Hence the vertex is at $$(3,0)$$ and the axis is the X-axis (rightwards).

For a parabola of the type $$y^{2}=4aX$$ the parametric coordinates of a general point are

$$X=a\,t^{2},\qquad Y=2a\,t,$$

where $$t$$ is the parameter. After shifting back to our actual variable $$x$$ (because $$X=x-3$$) we get

$$x-3=a\,t^{2},\qquad y=2a\,t.$$

Substituting $$a=4$$ gives the parametric point

$$x=4t^{2}+3,\qquad y=8t.$$

The next step is to write the tangent to the parabola at this parametric point. The standard tangent formula to $$y^{2}=4aX$$ is

$$t\,Y=X+a\,t^{2}.$$

Replacing $$X$$ by $$x-3$$ and $$Y$$ by $$y$$, and again taking $$a=4$$, we have

$$t\,y=(x-3)+4t^{2}.$$

This linear equation represents the tangent at the parameter value $$t$$. We rearrange it to slope-intercept form so that the slope is explicit:

$$$ t\,y = x-3+4t^{2} \;\Longrightarrow\; y=\frac{1}{t}\,x-\frac{3-4t^{2}}{t}. $$$

Thus the slope of this tangent is

$$m=\frac{1}{t}.$$

Let the external point from which we draw the tangents be $$P(h,k)$$. Because $$P$$ lies on each tangent, we substitute $$x=h,\;y=k$$ into the tangent equation:

$$$ t\,k = (h-3)+4t^{2}. $$$

Bringing all terms to one side gives a quadratic in $$t$$,

$$4t^{2}-k\,t+(h-3)=0.$$

Its two roots $$t_{1},t_{2}$$ correspond to the two tangents from $$P$$. For a quadratic $$at^{2}+bt+c=0$$ we know from Vieta’s formulae that

$$t_{1}+t_{2}=-\frac{b}{a},\qquad t_{1}t_{2}=\frac{c}{a}.$$

Here $$a=4,\;b=-k,\;c=h-3$$, so we obtain

$$t_{1}+t_{2}=\frac{k}{4},\qquad t_{1}t_{2}=\frac{h-3}{4}.$$

We are told that the two tangents are at right angles. If two lines with slopes $$m_{1},m_{2}$$ are perpendicular, the condition is $$m_{1}m_{2}=-1$$. We already noted that the slope of a tangent corresponding to parameter $$t$$ is $$m=\dfrac1t$$. Therefore

$$$ m_{1}m_{2}=\frac{1}{t_{1}}\cdot\frac{1}{t_{2}}=\frac{1}{t_{1}t_{2}}=-1. $$$

Hence

$$$ t_{1}t_{2}=-1. $$$

But we have also found that $$t_{1}t_{2}=\dfrac{h-3}{4}$$. Equating the two expressions for the product,

$$$ \frac{h-3}{4}=-1 \;\Longrightarrow\; h-3=-4 \;\Longrightarrow\; h=-1. $$$

The y-coordinate $$k$$ does not appear in this final condition, so the locus of $$P(h,k)$$ is the vertical straight line

$$x=-1,$$

which can be rewritten as $$x+1=0$$.

Hence, the correct answer is Option D.