Question

If the parabolas $ y^2 = 4x $ and $ x^2 = 32y $ intersect at $ (16, 8) $ at an angle $ \theta $ , then $ \theta $ equals to

• A. $ tan^{-1} 5/3 $
• B. $ tan^{-1} 4/5 $
• C. $ tan^{-1} 3/5 $
• D. $ \pi/2 $

Answer 1

Answer: (C)

Given curves are
$y^2 = 4x$
$\Rightarrow 2y \frac{dy}{dx} = 4 $
$\Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} = \frac{4}{16} $
$\Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} \frac{1}{4} = m_{1} $ (say)
and $x^{2} = 32y $
$ \Rightarrow 2x = 32 \frac{dy}{dx} $
$ \Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} = \frac{2\times16}{32} $
$\Rightarrow \left(\frac{dy}{dx}\right)_{\left(16,8\right)} = 1 = m_{2}$(say)
$ \therefore $ Angle between them, $\theta = tan^{-1}\left(\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right) $
$ = tan^{-1}\left(\frac{1-\frac{1}{4}}{1+1\times\frac{1}{4}}\right) $
$= tan^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)$
$ = tan^{-1}\left(\frac{3}{5}\right)$