Let T be the tangent to the parabola y^2 = 16x at the point (64,...

The equation of the parabola is $y^2 = 16x$. Comparing with $y^2 = 4ax$, we get $a = 4$.

Let the parametric coordinates of the point $(64, 32)$ be $(at_1^2, 2at_1)$.

$2at_1 = 32 \Rightarrow 8t_1 = 32 \Rightarrow t_1 = 4$

The slope of the tangent $T$ at $t_1$ is $m_1 = \dfrac{1}{t_1} = \dfrac{1}{4}$.

Let the tangent $L$ be at the point $(x_1, y_1)$ with parameter $t_2$. Its slope is $m_2 = \dfrac{1}{t_2}$.

Since $T$ and $L$ are perpendicular, $m_1 m_2 = -1$.

$\dfrac{1}{4} \times \dfrac{1}{t_2} = -1 \Rightarrow t_2 = -\dfrac{1}{4}$

The $x$-coordinate of the point $(x_1, y_1)$ is $x_1 = at_2^2 = 4\left(-\dfrac{1}{4}\right)^2 = \dfrac{1}{4}$.

The distance of a point $(x_1, y_1)$ on the parabola from the focus is given by its focal distance $x_1 + a$.

Distance $= \dfrac{1}{4} + 4 = \dfrac{17}{4}$

Answer: $\dfrac{17}{4}$