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

Solution:

Given:

Curves $y^{2}=8 x \ldots(1)$

$\& 2 x^{2}+y^{2}=10 \ldots(2)$

The point of intersection of two curves are $(0,0) \&(1,2 \sqrt{2})$

Now, Differentiating curves (1) \& (2) w.r.t x, we get

$\Rightarrow y^{2}=8 x$

$\Rightarrow 2 y \cdot \frac{d y}{d x}=8$

$\Rightarrow \frac{d y}{d x}=\frac{8}{2 y}$

$\Rightarrow \frac{d y}{d x}=\frac{4}{y} \ldots(3)$

$\Rightarrow 2 x^{2}+y^{2}=10$

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

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

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

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

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

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

$m_{1}=\frac{4}{y}$

$\Rightarrow \frac{4}{2 \sqrt{2}}$

$m_{1}=\sqrt{2} \ldots(5)$

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

$\Rightarrow \frac{-2 \times 1}{2 \sqrt{2}}$

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

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

Two curves intersect orthogonally if $m_{1} m_{2}=-1$, where $m_{1}$ and $m_{2}$ the slopes of the two curves.

$\Rightarrow \sqrt{2} \times \frac{-1}{\sqrt{2}}=-1$

$\therefore$ Two curves $y^{2}=8 x \& 2 x^{2}+y^{2}=10$ intersect orthogonally.