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)

$tan^{-1} 3/5$

Answer 2

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, $\theta = tan^{-1}\left(\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right)$
$= tan^{-1}\left(\frac{1-\frac{1}{4}}{1+1\times\frac{1}{4}}\right)$
$= tan^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)$
$= tan^{-1}\left(\frac{3}{5}\right)$