Question:
Write the angle between the curves $y^{2}=4 x$ and $x^{2}=2 y-3$ at the point $(1,2)$.
Solution:
Given two curves $y^{2}=4 x$ and $x^{2}=2 y-3$
Differentiating both the equations w.r.t. $x$,
$\Rightarrow 2 y \frac{d y}{d x}=4$ and $2 x=2 \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}=\frac{2}{y}$ and $\frac{d y}{d x}=x$
For $(1,2):$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{2}=1$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=1$
Thus we get
$\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$
$\Rightarrow \tan \theta=\left|\frac{1-1}{1+1}\right|$
$\Rightarrow \tan \theta=0$
$\Rightarrow \theta=0^{\circ}$