The equation of a tangent to the parabola $$y^2 = 8x$$ is $$y = x + 2$$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The standard form of the given parabola is $$y^{2}=4ax$$.

Comparing $$y^{2}=8x$$ with $$y^{2}=4ax$$ gives $$a=2$$.

For $$y^{2}=4ax$$ the tangent in slope form is

$$y = mx + \frac{a}{m} \qquad -(1)$$

1. Tangent already given
    The equation $$y = x + 2$$ has slope $$m_1 = 1$$.
    With $$m = 1$$, (1) gives $$c = \dfrac{a}{m}= \dfrac{2}{1}=2$$, so $$y = x + 2$$ is indeed a tangent.

2. Slope of the required second tangent
    The second tangent must be perpendicular to the first.
    If $$m_1 = 1$$, then $$m_2 = -\dfrac{1}{m_1} = -1$$.

3. Equation of the perpendicular tangent
    Put $$m = -1$$ in (1):

$$y = (-1)x + \frac{2}{-1} \; \Rightarrow \; y = -x - 2 \qquad -(2)$$

4. Point of intersection of the two tangents
    Solve $$y = x + 2$$ and $$y = -x - 2$$ simultaneously:

$$x + 2 = -x - 2 \;\Longrightarrow\; 2x = -4 \;\Longrightarrow\; x = -2$$

With $$x = -2$$, $$y = x + 2 = 0$$.

Thus the point common to both tangents (and lying on the given tangent) is $$(-2,0)$$.

Hence, the required point is $$(-2,0)$$.

Option D which is: $$(-2, 0)$$