If two tangents drawn from a point $$P$$ to the parabola $$y^2 = 4x$$ are at right angles, then the locus of $$P$$ is

Let the parabola be $$y^{2}=4x$$, i.e. $$a = 1$$.

Using the slope form, the equation of a tangent to $$y^{2}=4ax$$ with slope $$m$$ is
$$y = mx + \frac{a}{m} \;.$$

For our parabola $$a = 1$$, so every tangent can be written as
$$y = mx + \frac{1}{m}\;.\tag{-1}$$

Suppose a point $$P(h,k)$$ lies outside the curve and two tangents can be drawn from it. Because each tangent passes through $$P$$, substitute $$x = h,\; y = k$$ in $$-(1)$$:

$$k = mh + \frac{1}{m}\;.$$

Re-arrange into a quadratic in $$m$$:

$$mh - k + \frac{1}{m} = 0 \;\;\Longrightarrow\;\; h m^{2} - k m + 1 = 0\;.\tag{-2}$$

The roots $$m_{1}, m_{2}$$ of $$-(2)$$ are the slopes of the two tangents. For two lines to be perpendicular, the product of their slopes must equal $$-1$$:

$$m_{1}m_{2} = -1\;.\tag{-3}$$

From a quadratic $$Am^{2}+Bm+C=0$$, the product of its roots is $$\dfrac{C}{A}$$. For $$-(2)$$ we have $$A = h,\; C = 1$$, hence

$$m_{1}m_{2} = \frac{1}{h}\;.\tag{-4}$$

Equating $$-(3)$$ and $$-(4)$$:

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

Thus every point $$P(h,k)$$ from which the two tangents are at right angles must satisfy $$h = -1$$. Therefore the locus of $$P$$ is

$$x = -1\;.$$

Option B which is: $$x = -1$$