Question

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

Answer 1

We have

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

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

$4x+4-36\left(9-x\right)=0$

$40x+4-36\times 9=0$

$40x=36\times 9-4$

$10x=80$

$x=8$

Now,

Points of intersection are are:$(8,6)(8,-6)$

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

$\frac{dy}{dx}=\frac{2}{y}=\frac{2}{6}=\frac{1}{3}att=6$

$\frac{2}{-6}=\frac{-1}{3}aty=-6$

$2y\frac{dy}{dx}=-36$

$\frac{dy}{dx}=\frac{-18}{y}=-3aty=6$

$=3aty=-6$

$at\left(8,6\right)$

$\frac{1}{3}\times -3=-1$

$at\left(8,-6\right)$

$\frac{-1}{3}\times 3=-1$

Hence , curves intersection orthogonalliy