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
$8(x+4)=24(4-x)$
$x+4=12-3x$
$4x=8$
$\Rightarrow x=2$
$\therefore y^2=8(6)$
$\Rightarrow y=\pm 4\sqrt{3}$
For Eq. (i), $2y\frac{dy}{dx}=8$
$\Rightarrow 2y{m}_1=8$
$m_1=\frac{8}{2y}$
$\Rightarrow m_1=\frac{4}{y}=\frac{4}{4\sqrt{3}}=\frac{1}{\sqrt{3}}$
For Eq. (ii), $2y\frac{dy}{dx}=-24$
$\Rightarrow y{m}_2=-12$
$m_2=\frac{-12}{4\sqrt{3}}=-\sqrt{3}$
Here, $m_1{m}_2=\frac{1}{\sqrt{3}}\times -\sqrt{3}=-1$
So, angle between the curves $\tan \theta =\left(\frac{m_1-m_2}{1+m_1{m}_2}\right)$
$\Rightarrow \tan \theta =\frac{\frac{1}{\sqrt{3}}-(-\sqrt{3})}{1+(-1)}$
$\Rightarrow \tan \theta =\infty$
$\Rightarrow \tan \theta =\tan \frac{\pi }{2}$
$\Rightarrow \theta =\frac{\pi }{2}$