Question

The angle between the curve $y=e^{mx}$ and $y=e^{-mx}$ for $m>1$ is

• A. ${\tan }^{-1}\left(\frac{2m}{1-m^2}\right)$
• B. ${\tan }^{-1}\left(\frac{m}{1-m^2}\right)$
• C. ${\tan }^{-1}\left(\frac{-m}{1+m^2}\right)$
• D. ${\tan }^{-1}\left(\frac{2m}{m^2+1}\right)$

Answer 1

Answer: (A)

${\tan }^{-1}\left(\frac{2m}{1-m^2}\right)$

Given $y=e^{mx}$ and $y=e^{-mx}$

Multiply both $y^2=e^{mx}{e}^{-mx}=e^{mx-mx}=e^0=1$

$\Rightarrow y=1$, since $y>0$

So the point of intersection of both the curve is $(0,1)$

Now differentiate both the curve to get the slope of tangent at their point of intersection

$m_1=\frac{d}{dx}\left(e^{mx}\right){|}_{x=0}=m{e}^{mx}{|}_{x=0}=m$

And $m_2=\frac{d}{dx}\left(e^{-mx}\right){|}_{x=0}=-m{e}^{mx}{|}_{x=0}=-m$

If $\theta$ is the angle between them, then

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{2m}{1-m^2}\right|$

$\therefore \theta ={\tan }^{-1}\left|\frac{2m}{1-m^2}\right|$