#### Provide solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 2 sub question 3

$m_{1} \times m_{2}=-1$

Hence, two curves intersect orthogonally.

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

Given –   $x^{2}+4y^{2}=8---(1)$

$x^{2}-2y^{2}=4---(2)$

Solving (1) & (2),we get

From (2) curve,

$x^{2}=4+2y^{2}---(1)$

Substituting in (1)

\begin{aligned} &=x^{2}+4 y^{2}=8 \\ &=4+2 y^{2}+4 y^{2}=8 \\ &=6 y^{2}=4 \end{aligned}

\begin{aligned} &=y^{2}=\frac{4}{6} \\ &=y=\pm \sqrt{\frac{2}{3}} \end{aligned}

Substituting  $y=\pm \sqrt{\frac{2}{3}}$ in $x^{2}=4+2 y^{2}$ , we get

\begin{aligned} &=x^{2}=4+2\left(\pm \sqrt{\frac{2}{3}}\right)^{2} \\ &=x^{2}=4+2\left(\frac{2}{3}\right) \\ &=x^{2}=4+\frac{4}{3}=\frac{12+4}{3}=\frac{16}{3} \\ &=x=\pm \frac{4}{\sqrt{3}} \end{aligned}

The point of intersection of two curves is $\left(\frac{4}{\sqrt{3}}, \frac{2}{(\sqrt{3})}\right)\;\; \&\;\;\left(-\frac{4}{\sqrt{3}},-\frac{2}{(\sqrt{3})}\right)$

First curve is  $x^{2}+4 y^{2}=8$

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} &=2 x^{2}+8 y\left(\frac{d y}{d x}\right)=0 \\ &=8 y \frac{d y}{d x}=-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{8 y}=\frac{-x}{4 y} \\ &=m_{1}=\frac{d y}{d x}=\frac{-x}{4 y}---(3) \end{aligned}

Second curve is  $x^{2}-2 y^{2}=4$

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

\begin{aligned} &=\frac{d y}{d x}=\frac{2 x}{4 y}=\frac{x}{2 y} \\ &=m_{2}=\frac{d y}{d x}=\frac{x}{2 y}---(4) \end{aligned}

At $\left(\frac{4}{\sqrt{3}}, \frac{2}{(\sqrt{3})}\right)$ in eq (3), we get

\begin{aligned} &=\frac{d y}{d x}=\frac{\frac{-4}{\sqrt{3}}}{4 \times\left(\sqrt{\frac{2}{3}}\right)} \\ &=\frac{d y}{d x}=\frac{\frac{-1}{\sqrt{3}}}{\left(\sqrt{\frac{2}{3}}\right)} \end{aligned}

\begin{aligned} &=m_{1}=\frac{-1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{2}}=\left(\frac{-1}{\sqrt{2}}\right) \\ &=m_{1}=\frac{-1}{\sqrt{2}} \end{aligned}

At $\left(\frac{4}{\sqrt{3}}, \frac{2}{(\sqrt{3})}\right)$ in eq (4), we get

\begin{aligned} &=\frac{d y}{d x}=\frac{\frac{4}{\sqrt{3}}}{2 \times\left(\sqrt{\frac{2}{3}}\right)} \\ &=m_{2}=\frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{2 \sqrt{2}}=\left(\frac{2}{\sqrt{2}}\right)=\sqrt{2} \\ &=m_{2}=\sqrt{2} \end{aligned}

When $m_{1}=\frac{-1}{\sqrt{2}}$ and $m_{2}=\sqrt{2}$

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

$=\frac{-1}{\sqrt{2}} \times \sqrt{2}=-1$

Two curves  $x^{2}+4 y^{2}=8 \;\;\; \&\;\;\; x^{2}-2 y^{2}=4$ intersect orthogonally.