#### Explain solution rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 question 9

$m_{1} \times m_{2}=-1$ hence, the two curves intersect at right angles.

Hint –

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

Given –

$\begin{gathered} \frac{x^{2}}{a^{2}+d_{1}}+\frac{y^{2}}{b^{2}+d_{1}}=1---(1) \\ \quad \frac{x^{2}}{a^{2}+d_{2}}+\frac{y^{2}}{b^{2}+d_{2}}=1---(2) \end{gathered}$

Consider First curve is  $\begin{gathered} \frac{x^{2}}{a^{2}+d_{1}}+\frac{y^{2}}{b^{2}+d_{1}}=1 \\ \end{gathered}$

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}+d_{1}}+\frac{2 y}{b^{2}+d_{1}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}+d_{1}} \frac{d y}{d x}=\frac{-2 x}{a^{2}+d_{1}} \\ \end{aligned}

\begin{aligned} &=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{1}\right)}{y\left(a^{2}+d_{1}\right)}\\ &=m_{1}=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{1}\right)}{y\left(a^{2}+d_{1}\right)}---(3) \end{aligned}

Second curve is $\begin{gathered}\quad \frac{x^{2}}{a^{2}+d_{2}}+\frac{y^{2}}{b^{2}+d_{2}}=1 \end{gathered}$

Differentiating above with respect to x,
\begin{aligned} &=\frac{2 x}{a^{2}+d_{2}}+\frac{2 y}{b^{2}+d_{2}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}+d_{2}} \frac{d y}{d x}=\frac{-2 x}{a^{2}+d_{2}} \\ \end{aligned}

\begin{aligned} &=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{2}\right)}{y\left(a^{2}+d_{2}\right)} \\ &=m_{2}=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{2}\right)}{y\left(a^{2}+d_{2}\right)}---(4) \end{aligned}

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

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

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

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

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

From (3) & (4)

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

Now putting the value of  $\frac{x^{2}}{y^{2}}$ from eq (5) in eq (6)

Then,
\begin{aligned} &=\left\{\frac{\left(d_{2}-d_{1}\right)\left(a^{2}+d_{1}\right)\left(a^{2}+d_{2}\right)}{\left(b^{2}+d_{1}\right)\left(b^{2}+d_{2}\right)\left(d_{1}-d_{2}\right)}\right\}=\frac{\left(b^{2}+d_{1}\right)}{\left(a^{2}+d_{1}\right)} \frac{\left(b^{2}+d_{2}\right)}{\left(a^{2}+d_{2}\right)} \\ \end{aligned}

\begin{aligned} &=\frac{\left(d_{2}-d_{1}\right)}{\left(d_{1}-d_{2}\right)} \\ &=-\frac{\left(d_{1}-d_{2}\right)}{\left(d_{1}-d_{2}\right)} \\ &=-1 \end{aligned}

So, the curves intersect at right angles.

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