Question

Find the angle between the curves $y^2=4x+4$ and $y^2=36(9-x)$.

Answer 1

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

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

$=36\times 9-36x$

$40x=36\times 9-4$

$81\times 4-4$

$=80\times 4$

$x=8$

$y=\pm 6$

Intersection points $(8,\pm 6)$

$\frac{dy}{dx}$ for $y^2=4x+4$

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

$\frac{dy}{dx}$ for $y^2=36(9-x)$

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

$\frac{dy}{dx}=\frac{-18}{y}$

at $(8,6)$

$m_1=\frac{dy}{dx}{|}_{(8,6)}=\frac{2}{6}=\frac{1}{3}$

$m_2=\frac{dy}{dx}{|}_{(8,6)}=\frac{-18}{6}=-3$

as, $m_1{m}_2=-1$

$\theta ={90}^o$

Angle between curves is $\frac{\pi }{2}$.