Show that the following curves intersect orthogonally at the indicated points :
$x^{2}=y$ and $x^{3}+6 y=7$ at $(1,1)$
Given:
Curves $x^{2}=y \ldots$ (1)
$\& x^{3}+6 y=7 \ldots(2)$
The point of intersection of two curves $(1,1)$
Solving (1) \& (2), we get,
First curve is $x^{2}=y$
Differentiating above w.r.t $x$,
$\Rightarrow 2 x=\frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}=2 x$ .....(3)
Second curve is $x^{3}+6 y=7$
Differentiating above w.r.t $x$,
$\Rightarrow 3 x^{2}+6 \cdot \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{-3 x^{2}}{6}$
$\Rightarrow \frac{d y}{d x}=\frac{-x^{2}}{2}$
$\Rightarrow m_{2}=\frac{-x^{2}}{2} \ldots(4)$
Substituting $(1,1)$ for $m_{1} \& m_{2}$, we get,
$m_{1}=2 x$
$\Rightarrow 2 \times 1$
$m_{1}=2 \ldots(5)$
$m_{2}=\frac{-x^{2}}{2}$
$\Rightarrow \frac{-1^{2}}{2}$
$m_{2}=-\frac{-1}{2} \ldots(6)$
when $m_{1}=2 \& m_{2}=-\frac{-1}{2}$
$\Rightarrow 2 \times \frac{-1}{2}=-1$
$\therefore$ Two curves $x^{2}=y \& x^{3}+6 y=7$ intersect orthogonally.
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