If \(\theta\) denotes the acute angle between the curves \(y=10-x^2\) and \(y=2+x^2\), at a point of the intersection, then \(|\tan \theta|\) is equal to

\(\begin{aligned}
& y=10-x^2 ...(i)\\
& y=2+x^2...(ii)
\end{aligned}\)
Solving (i) and (ii), we get \(y=6\)
From (i), \(6=10-x^2 \Rightarrow x= \pm 2\) Differentiating (i) w.r.t. \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{(-2,6)}=4\)
Differentiating (ii) w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{(-2,6)}=-4 \\
\therefore \quad & \tan \theta=\left|\frac{4-(-4)}{1+4(-4)}\right|=\left|\frac{8}{-15}\right| \Rightarrow|\tan \theta|=\frac{8}{15}
\end{aligned}\)