Show that the following curves intersect orthogonally at the indicated points:

Solution:

Given:

Curves $x^{2}=4 y \ldots(1)$

$\& 4 y+x^{2}=8 \ldots(2)$

The point of intersection of two curves $(2,1)$

Solving $(1) \&(2)$, we get,

First curve is $x^{2}=4 y$c

Differentiating above w.r.t $\mathrm{x}$,

$\Rightarrow 2 x=4 \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x}{4}$

$\Rightarrow m_{1}=\frac{x}{2} \ldots(3)$

Second curve is $4 y+x^{2}=8$

$\Rightarrow 4 \cdot \frac{d y}{d x}+2 x=0$

$\Rightarrow \frac{d y}{d x}=\frac{-2 x}{4}$

$\Rightarrow m_{2}=\frac{-x}{2} \ldots(4)$

Substituting $(2,1)$ for $m_{1} \& m_{2}$, we get,

$m_{1}=\frac{x}{2}$

$\Rightarrow \frac{2}{2}$

$m_{1}=1 \ldots(5)$

$m_{2}=\frac{-x}{2}$

$\Rightarrow \frac{-2}{2}$

$m_{2}=-1 \ldots(6)$

when $m_{1}=1 \& m_{2}=-1$

$\Rightarrow 1 \times-1=-1$

$\therefore$ Two curves $x^{2}=4 y \& 4 y+x^{2}=8$ intersect orthogonally.