The angle between the curves $y^2=4x+4$ and $y^2=36\left(9-x\right)$

$y^2=4\left(x+1\right),y^2=36\left(9-x\right)$ 
Point of intersection of curves are $\left(8,6\right),\left(8,-6\right)$                  
From $y^2=4\left(x+1\right)$
$m_1=\frac{2}{y}$
from$y^2=36(9-x)$
$m_2=\frac{-18}{y}$ 
at p(8, 6) 
$m_1=\frac{2}{6}=\frac{1}{3}$ 
$m_2=\frac{-18}{6}=-3$

$m_1\times m_2=\frac{1}{3}\times -3=-1$ 
 $m_1{m}_2=-1$

So given curves are Orthogonal,angle between them  $\theta ={90}^0$