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