The angle between \(y^{2}=4 x\) and \(x^{2}+y^{2}=12\) at a point of their intersection is

Given equation of curve is
and
\[
\begin{gathered}
y^{2}=4 x \\
x^{2}+y^{2}=12
\end{gathered}
\]
Let the slope of curve (i) is \(m_{1}\) and curve (ii) is \(\left(m_{2}\right)\).
Then, from Eqs. (i) and (ii),
\[
2 y \frac{d y}{d x}=4 \Rightarrow m_{1}=\frac{2}{y}
\]
and
\[
\Rightarrow \quad m_{2}=\frac{-x}{y}
\]
Let the angle between curve at intersection point is ' \(\theta\).
\[
\begin{aligned}
\tan \theta &=\left|\frac{m_{1}-m_{2}}{1+m_{1} \cdot m_{2}}\right|=\left|\frac{2 / y-(-x / y)}{1+(2 / y)(-x / y)}\right| \\
\tan \theta &=\left|\frac{\frac{2}{y}+\frac{x}{y}}{1-\frac{2 x}{y^{2}}}\right|=\left|\frac{(2+x) y}{y^{2}-2 x}\right| \quad \ldots \text { (iii) }
\end{aligned}
\]
On solving Eqs. (i) and (ii), we get
\[
\begin{aligned}
x^{2}+4 x=12 \\
\Rightarrow & x^{2}+4 x-12=0 \\
\Rightarrow & x^{2}+6 x-2 x-12=0 \\
\Rightarrow &(x+6)(x-2)=0 \\
\Rightarrow &x=-6,2 \quad \text { (here } x \neq-6) \\
\Rightarrow & y=\pm 2 \sqrt{2}
\end{aligned}
\]
So, the intersection point \(=(2 \pm 2 \sqrt{2})\)
From Eq. (iii)
\[
\begin{gathered}
\tan \theta=\left|\frac{(2+2)(\pm 2 \sqrt{2})}{8-4}\right|=\left|4 \frac{(\pm 2 \sqrt{2})}{4}\right| \\
\tan \theta=|\pm 2 \sqrt{2}|=2 \sqrt{2} \Rightarrow \theta=\tan ^{-1}(2 \sqrt{2})
\end{gathered}
\]