#### Explain solution rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 question 8 sub question 1

$b^{2}=a^{2}$

Hint –

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

Given –

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

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

Differentiating above with respect to x,

As we know,$\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{2 x}{a^{2}} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{2 x}{a^{2}} \times \frac{b^{2}}{2 y} \\ &=\frac{d y}{d x}=\frac{b^{2} x}{a^{2} y} \\ &=m_{1}=\frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}---(3) \end{aligned}

Second curve is $xy=c^{2}$

Differentiating above with respect to x,
\begin{aligned} &=x \frac{d y}{d x}+y=0 \\ &=\frac{d y}{d x}=\frac{-y}{x} \\ &=m_{2}=\frac{d y}{d x}=\frac{-y}{x}---(4) \end{aligned}

Two curves intersects orthogonally if  $m_{1}$ & $m_{2}$ =-1

Since  $m_{1}$ & $m_{2}$ cuts orthogonally
\begin{aligned} &=\frac{b^{2} x}{a^{2} y} \times \frac{-y}{x}=-1 \\ &=b^{2}=a^{2} \end{aligned}