Question

The angle between the curves $y^{2}=8(x+4)$ and $y^{2}=24(4-x)$ is

• A. $\tan ^{-1}\left(\frac{1}{6}\right)$
• B. $\tan ^{-1}(3)$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (C)

Given, curves are
$y^{2}=8(x+4)\,\,\,...(i)$
and $y^{2}=24(4-x)\,\,\,...(ii)$
By Eqs.(i) and (ii), we get
$ 8(x+4) =24(4-x) $
$x+4 =12-3 x $
$4 x =8$
$\Rightarrow x=2 $
$\therefore y^{2} =8(6) $
$\Rightarrow y=\pm 4 \sqrt{3} $
For Eq. (i), $2 y \frac{d y}{d x}=8 $
$\Rightarrow 2 y m_{1}=8$
$m_{1}=\frac{8}{2 y} $
$\Rightarrow m_{1}=\frac{4}{y}=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}$
For Eq. (ii), $2 y \frac{d y}{d x}=-24 $
$\Rightarrow y \,m_{2}=-12$
$m_{2}=\frac{-12}{4 \sqrt{3}}=-\sqrt{3}$
Here, $m_{1} \,m_{2}=\frac{1}{\sqrt{3}} \times-\sqrt{3}=-1$
So, angle between the curves $\tan \theta=\left(\frac{m_{1}-m_{2}}{1+m_{1} \,m_{2}}\right)$
$\Rightarrow \tan \theta=\frac{\frac{1}{\sqrt{3}}-(-\sqrt{3})}{1+(-1)} $
$\Rightarrow \tan \theta=\infty$
$\Rightarrow \tan \theta=\tan \frac{\pi}{2} $
$\Rightarrow \theta=\frac{\pi}{2}$