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: , 2
Since, the equation of curves are
$y=10-x^{2}$........(1)
$y=2+x^{2}$.........(2)
Adding eqn (1) and (2), we get
$2 y=12 \Rightarrow y=6$
Then, from eqn (1)
$x=\pm 2$
Differentiate equation (1) with respect to $x$
$\frac{d y}{d x}=-2 x \Rightarrow\left(\frac{d y}{d x}\right)_{(2,6)}=-4$ and $\left(\frac{d y}{d x}\right)_{(-2,6)}=4$
Differentiate equation (2) with respect to $x$
$\frac{d y}{d x}=2 x \Rightarrow\left(\frac{d y}{d x}\right)_{(2,6)}=4$ and $\left(\frac{d y}{d x}\right)_{(-2,6)}=-4$
At $(2,6) \tan \theta=\left(\frac{(-4)-(4)}{1+(-4) \times(4)}\right)=\frac{-8}{15}$
At $(-2,6), \tan \theta=\frac{(4)-(-4)}{1+(4)(-4)}=\frac{8}{-15}$
$\therefore|\tan \theta|=\frac{8}{15}$