Find the angle between the curves $y^2=8x, 4x^2+y^2=32$

Given curves 

$y^2=8x\to \left(1\right);$

$4x^2+y^2=32\to \left(2\right)$

from (2) $4x^2+y^2-32=0$

$x^2+2x-8=0\Rightarrow x^2+4x-2x-8=0$

$x(x+4)-2(x+4)=0$

$x+4=0 x-2=0$

$x=-4 x=2$

from (1) $y^2=8x$

if $x=-4\Rightarrow y^2=-32$(does not satisfying)

if $x=2\Rightarrow y^2=8\left(2\right)$

$y=\pm 4\Rightarrow \therefore P=(2,4),Q=(2,-4)$

from (1) $y^2=8x$

diff w.r.t ‘$x$’

$2y\frac{dy}{dx}=8\Rightarrow \frac{dy}{dx}=\frac{8}{2y}=\frac{4}{y}$

slope $={\left(\frac{dy}{dx}\right)}_{P(2,4)}\Rightarrow m_1=\frac{4}{4}=1$

from (2) $4x^2+y^2=32$

diff w.r.t ‘$x$’

$4\left(2x\right)+2y\frac{dy}{dx}=0\Rightarrow 2y\frac{dy}{dx}=-8x$

$\frac{dy}{dx}=\frac{-8x}{2y}=\frac{-4x}{y}$

slope $\left(m_2\right)={\left(\frac{dy}{dx}\right)}_{P(2,4)}\Rightarrow m_2=\frac{-4\left(2\right)}{4}=-2$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{1+2}{1-2}\right|\Rightarrow \tan \theta =\left|\frac{3}{-1}\right|$

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