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  • Explain solution rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 question 8 subquestion 2

Explain solution rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 question 8 subquestion 2

Answers (1)

a^{2}-b^{2}=A^{2}+B^{2}Hint –

Two curves intersects orthogonally if m_{1}m_{2}=-1, where m1 and m2 are the slopes of two curves.

Given –

\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1---(1) \\ &\quad \frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1---(2) \end{aligned}

 Consider First curve  \begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \end{aligned}

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} 

Consider second curve \begin{aligned} \frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1 \end{aligned}

Differentiating above with respect to x,

\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_{2}=\frac{d y}{d x}=\frac{B^{2} x}{A^{2} y}---(4) \end{aligned}

Two curves intersects orthogonally if  m_{1}\times m_{2}=-1

\begin{aligned} &=\frac{-b^{2} x}{a^{2} y} \times \frac{B^{2} x}{A^{2} y}=-1\\ &=\frac{x^{2} b^{2} B^{2}}{y^{2} a^{2} A^{2}}=1\\ &=\frac{x^{2}}{y^{2}}=\frac{\left(a^{2} A^{2}\right)}{\left(b^{2} B^{2}\right)}---(5) \end{aligned}

 

Now, subtract eq (1) & (2), we get

 

\begin{aligned} &=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1-1 \\ &=x^{2}\left(\frac{1}{a^{2}}-\frac{1}{A^{2}}\right)+y^{2}\left(\frac{1}{b^{2}}+\frac{1}{B^{2}}\right)=0 \\ \end{aligned}

\begin{aligned} &=x^{2}\left(\frac{A^{2}-a^{2}}{a^{2} A^{2}}\right)+y^{2}\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right)=0 \\ &=x^{2}\left(\frac{A^{2}-a^{2}}{a^{2} A^{2}}\right)=-y^{2}\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \end{aligned}

\begin{aligned} &=\frac{x^{2}}{y^{2}}=-\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \times\left(\frac{a^{2} A^{2}}{A^{2}-a^{2}}\right) \\ &=\frac{x^{2}}{y^{2}}=-\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \times(-1)\left(\frac{a^{2} A^{2}}{a^{2}-A^{2}}\right) \\ &=\frac{x^{2}}{y^{2}}=-\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \times\left(\frac{a^{2} A^{2}}{a^{2}-A^{2}}\right)---(6) \end{aligned}

Put value of  \frac{x^{2}}{y^{2}} in eq (5), we get
\begin{aligned} &=\left(\frac{a^{2} A^{2}}{a^{2}-A^{2}}\right)\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right)=\frac{\left(a^{2} A^{2}\right)}{\left(b^{2} B^{2}\right)} \\ &=\left(\frac{B^{2}+b^{2}}{a^{2}-A^{2}}\right)=1 \\ &=B^{2}+b^{2}=a^{2}-A^{2} \\ &=a^{2}-b^{2}=A^{2}+B^{2} \end{aligned}

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