$\begin{array}{1 1}(A)\;\tan^{-1} (\frac{3}{5}) \\(B)\;\tan^{-1} (\frac{4}{5}) \\(C)\;\pi \\(D)\;\frac{\pi}{2} \end{array}$

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The slope of the tangent to $y^2=4x$ at $(16,8) $ is given by

$m_1 =\bigg( \large\frac{dy}{dx}\bigg)_{(16,8)}=\bigg( \large\frac{4}{2y} \bigg)_{(16,8)}$

$\qquad= \large\frac{2}{8}$

$\qquad=\large\frac{1}{4}$

The slope of the tangent to $x^2=32y$ at $(16,8)$ is given by

$m_2= \bigg(\large\frac{dy}{dx}\bigg)_{16,8}$

$\qquad= \bigg(\large\frac{2x}{32}\bigg)_{16,8}$

$\qquad=1$

$\tan \theta=\large\frac{1- \frac{1}{4}}{1+\frac{1}{4}}$

$\qquad= \large\frac{3}{5}$

$\theta= \tan ^{-1} \bigg(\large\frac{3}{5} \bigg)$

Hence A is the correct answer.

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