Question

The angel of intersection between the curves $y^2=4ax$ and $x^2=32y$ at point $(16,8)$ is

• A. $\pi$
• B. $ta{n}^{-1}\left(\frac{4}{5}\right)$
• C. $ta{n}^{-1}\left(\frac{3}{5}\right)$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $ta{n}^{-1}\left(\frac{3}{5}\right)$

$y^2=4ax$

$2y\frac{dy}{dx}=4a$

or, $\frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}$

$\therefore \frac{dy}{dx}{|}_{16,8}=\frac{2a}{8}=\frac{a}{4}$

$x^2=32y$

$2x=32\frac{dy}{dx}$

or, $\frac{dy}{dx}=\frac{x}{16}$

$\therefore \frac{dy}{dx}{|}_{16,8}=\frac{16}{16}=1$

Now, $\tan \theta =\frac{w_1-w_2}{1+w_1{w}_2}=\left|\frac{\frac{a}{4}-1}{1-(\frac{a}{4}\times 1)}\right|=\left|\frac{a-4}{4+a}\right|$

Now, according to option, value of $a=1$

So, $\tan \theta =\left|\frac{1-4}{4+1}\right|=\frac{3}{5}\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{5}\right)$ is the required angle.