Find the angle between the curves $x^2+3y=3$ and $x^2-y^2+25=0$

Given 

$x^2+3y=3........\left(1\right)$

$x^2-y^2+25=0........\left(2\right)$

$x^2=3-3y$

$x^2$ value sub in equation (2)

$3-3y-y^2+25=0-y^2-3y+28=0$

$y^2+3y-28=0;y^2+7y-4y-28=0$

$y(y+7)-4(y+7)=0;(y+7)(y-4)=0$

$y=-7,4;\therefore x^2=3-3y$

if $y=-7\Rightarrow x^2=3+21$

$x^2=24\Rightarrow x=\sqrt{24}=2\sqrt{6}$

$\therefore P=(2\sqrt{6},-7)$

if $y = 4$

if $y = 4\Rightarrow x^2=3-3\left(4\right)=3-12\Rightarrow x^2\ne -9$

from (1) $x^2+3y=3$

diff w.r.t ‘$x$’

$2x+3\frac{dy}{dx}=0\Rightarrow 3\frac{dy}{dx}=-2x\Rightarrow \frac{dy}{dx}=\frac{-2x}{3}$

slope $\left(m\right)={\left(\frac{dy}{dx}\right)}_{P(2\sqrt{6},-1)}$

$m_1=\frac{-2\left(2\sqrt{6}\right)}{3}{m}_1=\frac{-4\sqrt{6}}{3}$

from (1) $x^2-y^2+25+0$

diff w.r.t ‘$x$’

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

slope $={\left(\frac{dy}{dx}\right)}_{P\left(x\sqrt{6}\right)}$

$m_2=\frac{2\sqrt{6}}{-7};m_2=\frac{-2\sqrt{6}}{7}$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{-4\sqrt{6}}{3}+\frac{2\sqrt{6}}{7}}{1+\left(\frac{4\sqrt{6}}{3}\right)\left(\frac{-2\sqrt{6}}{7}\right)}\right|$

$=\left|\frac{-28\sqrt{6}+6\sqrt{6}}{21+48}\right|=\left|\frac{-22\sqrt{6}}{69}\right|$

$\theta ={\tan }^{-1}\left(\frac{22\sqrt{6}}{69}\right)$