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

Answers (1)


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}

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