Find the condition for the following set of curves to interest orthogonally.

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Question:

Find the condition for the following set of curves to interest orthogonally.

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $x y=c^{2}$

Solution:

Given:

Curves $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ .....(1)

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

First curve is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Differentiating above w.r.t $x$,

$\Rightarrow \frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{y}{b^{2}} \cdot \frac{d y}{d x}=\frac{x}{a^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{\frac{x}{y^{2}}}{b^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}$

$\Rightarrow m_{1}=\frac{b^{2} x}{a^{2} y} \ldots(3)$

Second curve is $x y=c^{2}$

$\Rightarrow \Rightarrow 1 \times y+x . \frac{d y}{d x}=0$

$\Rightarrow x . \frac{d y}{d x}=-y$

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

When $m_{1}=\frac{b^{2} x}{a^{2} y} \& m_{2}=\frac{-y}{x}$

Since ,two curves intersect orthogonally,

$\Rightarrow \frac{-b^{2} x}{a^{2} y} \times \frac{-y}{x}=-1$

$\Rightarrow \frac{b^{2}}{a^{2}}=1$

$\Rightarrow \therefore a^{2}=b^{2}$

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