Question:
If $\theta$ denotes the acute angle between the curves, $y=10-x^{2}$ and $y=2+x^{2}$ at a point of their intersection, then $|\tan \theta|$ is equal to :
Correct Option: , 4
Solution:
Point of intersection is $\mathrm{P}(2,6)$.
Also, $\mathrm{m}_{1}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{P}(2,6)}=-2 \mathrm{x}=-4$
$\mathrm{m}_{2}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{P}(2,6)}=2 \mathrm{x}=4$
$\therefore|\tan \theta|=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\frac{8}{15}$
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