Show that the curves $y^2=4(x+1)$ and $y^2=36(9-x)$ intersect orthogonally.

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

$y^2=36(9-x)\to \left(2\right)$

From (1) & (2),

$4(x+1)=36(9-x)$

$x+1=9(9-x)\Rightarrow x+1=81-9x$

$10x=80\Rightarrow x=8$ sub in (1)

$y^2=4(8+1)=4\left(9\right)=36\Rightarrow y=\pm 6$

the pts of inter sections of curves

$p(8,6) \mathrm{\&} (8,-6)$

At $p(8,6)$

$y^2=4(x+1);y^2=36(9-x)$

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

$m_1{m}_2=-1\Rightarrow \theta ={90}^{\circ }$

similarly at $p(8,-6); m_1{m}_2=-1$

$\therefore$ Given curves intersect orthogonally