Question

If $\theta$ is angle of intersection between $y=10-x^2$ and $y=4+x^2$ then $|\tan \theta |$ is-

• A. $\frac{5\sqrt{3}}{11}$
• B. $\frac{7\sqrt{3}}{15}$
• C. $\frac{4\sqrt{3}}{11}$
• D. $none$

Answer 1

The correct option is C $\frac{4\sqrt{3}}{11}$

Given:$y=10-x^2$ ......$\left(1\right)$

$y=4+x^2$ ......$\left(2\right)$

On solving,

$10-x^2=4+x^2$

$\Rightarrow 2x^2=6$

$\Rightarrow x^2=\frac{6}{2}=3$

$\therefore x=\pm \sqrt{3}$ from $\left(1\right)$

$\Rightarrow y=10-{(\pm \sqrt{3})}^2=10-3=7$

Consider $y=10-x^2$

$\frac{dy}{dx}=-2x$

$m_1={\left[\frac{dy}{dx}\right]}_{(\sqrt{3},7)}=-2\times \sqrt{3}=-2\sqrt{3}$

Consider $y=4+x^2$

$\frac{dy}{dx}=2x$

$m_2={\left[\frac{dy}{dx}\right]}_{(\sqrt{3},7)}=2\times \sqrt{3}=2\sqrt{3}$

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

$\tan \theta =\left|\frac{-2\sqrt{3}-2\sqrt{3}}{1+(-2\sqrt{3})\left(2\sqrt{3}\right)}\right|$

$=\left|\frac{-4\sqrt{3}}{1-12}\right|$

$=\frac{4\sqrt{3}}{11}$

$\therefore \tan \theta =\frac{4\sqrt{3}}{11}$