Parabola

Concept:Use the standard tangent equation of a parabola and the condition for perpendicular slopes to find the unknown point, then compute its distance from the focus.Explanation:The parabola is $y^2 = 16x$. Compare with $y^2 = 4ax$ to get $4a = 16 \Rightarrow a = 4$.Focus is at $(a,0) = (4,0)$.Tangent $T$ is at $(64,32)$. For $y^2 = 4ax$, tangent at $(x_1,y_1)$ is $yy_1 = 2a(x+x_1)$.At $(64,32)$: $y \cdot 32 = 2 \cdot 4 (x+64) \Rightarrow 32y = 8x + 512 \Rightarrow x - 4y + 64 = 0$.Slope $m_T = \frac{1}{4}$ (from $y = \frac14 x + 16$).Since $L$ is perpendicular to $T$, slope $m_L = -4$.For a point $(x_1,y_1)$ on the parabola, slope of tangent is $\frac{2a}{y_1} = \frac{8}{y_1}$.Set $\frac{8}{y_1} = -4 \Rightarrow y_1 = -2$.Because it lies on $y^2 = 16x$, $(-2)^2 = 16x_1 \Rightarrow 4 = 16x_1 \Rightarrow x_1 = \frac14$.The point is $\left(\frac14, -2\right)$.Distance from focus $(4,0)$: $\sqrt{\left(4-\frac14\right)^2 + (0+2)^2} = \sqrt{\left(\frac{15}{4}\right)^2 + 4} = \sqrt{\frac{225}{16} + \frac{64}{16}} = \sqrt{\frac{289}{16}} = \frac{17}{4}$.Answer:$\boxed{\frac{17}{4}}$ (Option C).