Question

Let $\theta$ be the acute angle between the curves $y=|x^2-2|$ and $y=\sqrt{8-x^2}$ at their point(s) of intersection. Then the value of $\tan \theta$ is

• A. $0$
• B. $\frac{3}{5}$
• C. $\frac{5}{3}$
• D. $1$

Answer 1

The correct option is C $\frac{5}{3}$

$y=|x^2-2|$ and $y=\sqrt{8-x^2}$
Points of intersection:
$|x^2-2|=\sqrt{8-x^2}$
$\Rightarrow x^4-4x^2+4=8-x^2\Rightarrow x^4-3x^2-4=0\Rightarrow (x^2+1)(x^2-4)=0\Rightarrow x^2=4\Rightarrow x=\pm 2$
$\therefore y=2$

At $x=2$,
$y=x^2-2\Rightarrow \frac{dy}{dx}=2x{m}_1={\frac{dy}{dx}|}_{(2,2)}=4y=\sqrt{8-x^2}\Rightarrow \frac{dy}{dx}=\frac{-x}{y}{m}_2={\frac{dy}{dx}|}_{(2,2)}=-1\Rightarrow \tan \theta =\left|\frac{4-(-1)}{1-4}\right|=\frac{5}{3}$