Question

Find the abscissa of point $P$ and $Q$ on the curve $y=e^x+e^{-x}$ such that tangents at $P$ and $Q$ make ${60}^o$ with $x-axis$.

Answer 1

Slope at point $P$ is

$m_1=\tan {60}^o=\sqrt{3}$

Slope at point $Q$ is

$m_2=\tan {60}^o=\sqrt{3}$

$\frac{dy}{dx}=e^x-e^{-x}=\tan {60}^o=\sqrt{3}$

$\Rightarrow$ $e^x-\frac{1}{e^x}=\sqrt{3}$

$\Rightarrow$ $e^{2x}-1=\sqrt{3}{e}^x$

$\Rightarrow$ $e^{2x}-\sqrt{3}{e}^x-1=0$

let $e^x=t$ we get,

$\Rightarrow$ $t^2-\sqrt{3}t-1=0$

$t=\frac{\sqrt{3}\pm \sqrt{3+4}}{2}$

$t=\frac{\sqrt{3}\pm \sqrt{7}}{2}$

$\Rightarrow$ $e^x=\frac{\sqrt{3}\pm \sqrt{7}}{2}$

$\Rightarrow$ $x=ln\left(\frac{\sqrt{3}\pm \sqrt{7}}{2}\right)$

We know that value under $ln$ must be greater than $0$.

$\therefore$ $x=ln\left(\frac{\sqrt{3}+\sqrt{7}}{2}\right)$