Question 2

For the parabola $y^2 = 16x$, $4a = 16 \implies a = 4$. The focus is $S(4, 0)$. The slope of the tangent at $(x_0, y_0)$ is given by $\frac{dy}{dx} = \frac{8}{y}$. For tangent $T$ at $(64, 32)$, slope $m_T = \frac{8}{32} = \frac{1}{4}$. Since tangent $L$ is perpendicular to $T$, its slope $m_L = -4$. For a parabola $y^2 = 4ax$, the point of tangency with slope $m$ is $(\frac{a}{m^2}, \frac{2a}{m})$. So, $(x_1, y_1) = (\frac{4}{(-4)^2}, \frac{2(4)}{-4}) = (\frac{4}{16}, -2) = (\frac{1}{4}, -2)$. The distance of any point $(x_1, y_1)$ on the parabola from the focus is the focal distance $x_1 + a$. Distance $= \frac{1}{4} + 4 = \frac{17}{4}$.