Show that the following set of curves intersect orthogonally:

Solution:

Given:

Curves $y=x^{3} \ldots(1)$

$\& 6 y=7-x^{2} \ldots(2)$

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

$\Rightarrow 6 y=7-x^{2}$

$\Rightarrow 6\left(x^{3}\right)=7-x^{2}$

$\Rightarrow 6 x^{3}+x^{2}-7=0$

Since $f(x)=6 x^{3}+x^{2}-7$

we have to find $f(x)=0$, so that $x$ is a factor of $f(x)$.

when $x=1$

$f(1)=6(1)^{3}+(1)^{2}-7$

$f(1)=6+1-7$

$f(1)=0$

Hence, $x=1$ is a factor of $f(x)$.

Substituting $x=1$ in $y=x^{3}$, we get

$y=1^{3}$

$y=1$

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

First curve $y=x^{3}$

Differentiating above w.r.t $x$,

$\Rightarrow m_{1}=\frac{d y}{d x}=3 x^{2}$         .....(3)

Second curve $6 y=7-x^{2}$

Differentiating above w.r.t $x$,

$\Rightarrow 6 \frac{\mathrm{dy}}{\mathrm{dx}}=0-2 \mathrm{x}$

$\Rightarrow \mathrm{m}_{2}=\frac{-2 \mathrm{x}}{6}$

$\Rightarrow \mathrm{m}_{2}=\frac{-\mathrm{x}}{3} \ldots(4)$

At $(1,1)$, we have,

$m_{1}=3 x^{2}$

$\Rightarrow 3 \times(1)^{2}$

$\mathrm{m}_{1}=3$

At $(1,1)$, we have,

$\Rightarrow m_{2}=\frac{-x}{3}$

$\Rightarrow \frac{-1}{3}$

$\Rightarrow m_{2}=\frac{-1}{3}$

When $m_{1}=3 \& m_{2}=\frac{-1}{3}$

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

$\therefore$ Two curves $y=x^{3} \& 6 y=7-x^{2}$ intersect orthogonally.