Question

$\mathrm{If}\mathrm{the}\mathrm{parabolas}{y}^2=4x\mathrm{and}{x}^2=32y\mathrm{intersect}\mathrm{at}(16,8)\mathrm{at}\mathrm{an}\mathrm{angle}\theta ,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\theta \mathrm{is}$ .

• A. ${\tan }^{-1}\left(\frac{3}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{5}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{4}{5}\right)$
• D. ${\tan }^{-1}\left(\frac{5}{4}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{3}{5}\right)$

The slope of the tangent to $y^2$=4x at (16,8) is given by
$m_1=(\frac{\mathrm{dy}}{\mathrm{dx}}{)}_{(16,8)}=(\frac{4}{2y}{)}_{(16,8)}=\frac{1}{4}$
The slope of the tangent to $x^2$=32y at (16,8) is given by
$m_2=(\frac{\mathrm{dy}}{\mathrm{dx}}{)}_{(16,8)}=(\frac{2x}{32}{)}_{(16,8)}=1$
$\therefore \tan \theta =\left(\frac{1-(\frac{1}{4})}{1+\left(\frac{1}{4}\right)}\right|=\frac{3}{5}\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{5}\right)\left(\mathrm{acute}\mathrm{angle}\right)\Rightarrow \mathrm{or}\theta =\pi -{\tan }^{-1}\left(\frac{3}{5}\right)\left(\mathrm{obtuse}\mathrm{angle}\right)$