Find the angle between the curves $y^2=4x$ and $x^2+y^2=5$

Given curves $y^2=4x......\left(1\right)$

$x^2+y^2=5.......\left(2\right)$

Sub (1) in (2) $x^2+4x=5$

$\Rightarrow x^2+4x-5=0\Rightarrow x^2+5x-x-5=0$

$\Rightarrow x(x+5)-1(x+5)=0$

$\Rightarrow (x-1)(x+5)=0$

$x=1,-5$

If $x=1$, then $\left(1\right)\Rightarrow y^2=4\left(1\right)\Rightarrow y=\pm 2$

$\therefore$ intersection pt’s  $P(1,2),Q(1,-2)$

If $x=– 5$ then $\text{ (1) }\Rightarrow y^2=4(-5)=-20$

$y=\sqrt{-20}$ is not possible

$\text{ At }P(1,2)\text{ : }$

$y^2 = 4x x^2 + y^2 = 5$

$\mathrm{diff} \mathrm{w}.\mathrm{r}. \mathrm{to} ‘x’ \mathrm{diff} \mathrm{w}.\mathrm{r}. \mathrm{to} ‘x’$

$\Rightarrow 2y\frac{dy}{dx}=4x \Rightarrow 2x+2y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{4x}{2y}=\frac{2x}{y} \Rightarrow 2\mathrm{y}\frac{dy}{dx}=-2x$

Now $m_1={\left(\frac{dy}{dx}\right)}_{P(1,2)}=\frac{2}{2}=1\Rightarrow \frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}$

$m_2={\left(\frac{dy}{dx}\right)}_{p(1,2)}=-\frac{1}{2}$

Let '$\theta$ ' be the angle b/w (1) and (2)

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{1+\frac{1}{2}}{1+\left(1\right)\left(\frac{-1}{2}\right)}\right|=\left|\frac{3/2}{1-\frac{1}{2}}\right|=\left|\frac{3/2}{1/2}\right|=3$

$\theta ={\tan }^{-1}\left(3\right)$

Similarly the angle between the curves at the other point is also same.