The equation of the normal to the parabola, $$x^2 = 8y$$ at $$x = 4$$ is

The parabola is $$x^2 = 8y$$, i.e. $$y = \frac{x^2}{8}$$.

For any differentiable curve, the slope of the tangent at a point is $$\frac{dy}{dx}$$. Differentiate:

$$2x = 8\frac{dy}{dx} \;\; \Longrightarrow \;\; \frac{dy}{dx} = \frac{x}{4}$$

At the specified abscissa $$x = 4$$, the ordinate is

$$y = \frac{4^2}{8} = \frac{16}{8} = 2$$

Thus the point of tangency (and normality) is $$(4,\,2)$$. The tangent slope there is

$$m_{\text{tangent}} = \left.\frac{dy}{dx}\right|_{x=4} = \frac{4}{4} = 1$$

The normal is perpendicular to the tangent, so

$$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -1$$

Using point-slope form through $$(4,2)$$:

$$y - 2 = -1\,(x - 4)$$

Simplify:

$$y - 2 = -x + 4 \;\; \Longrightarrow \;\; x + y = 6$$

Therefore, the required normal is Option D which is: $$x + y = 6$$.