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

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

Hence, two curves intersect orthogonally.

Hint - Two curves intersect orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given –  \begin{aligned} x^{3}-3 x y^{2}=-2---(1) \\ \end{aligned}

$3 x^{2} y-y^{3}=2---(2)$

Adding (1) & (2), we get

\begin{aligned} &=x^{3}-3 x y^{2}+3 x^{2} y-y^{3}=-2+2 \\ &=x^{3}-3 x y^{2}+3 x^{2} y-y^{3}=0 \\ &=(x-y)^{3}=0 \\ &=(x-y)=0 \\ &=x=y \end{aligned}

Substituting  x=y in eq(1), we get

\begin{aligned} &=x^{3}-3 x(x)^{2}=-2 \\ &=x^{3}-3 x^{3}=-2 \\ &=-2 x^{3}=-2 \\ &=x^{3}=1 \\ &=x=1 \end{aligned}

Since x=y    (y=1)

The point of intersection of two curves is (1, 1)

First curve is  $x^{3}-3 x y^{2}=-2$

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} &=3 x^{2}-3\left(1 \times y^{2}-x \times 2 y \frac{d y}{d x}\right)=0 \\ &=3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \end{aligned}

\begin{aligned} &=3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{3\left(x^{2}-y^{2}\right)}{6 x y} \\ &=m_{1}=\frac{d y}{d x}=\frac{\left(x^{2}-y^{2}\right)}{2 x y}---(3) \end{aligned}

Second curve is  $3 x^{2} y-y^{3}=2$

Differentiating above with respect to x,

\begin{aligned} &=3\left(2 x y+\frac{x^{2} d y}{d x}\right)-3 y^{2} \frac{d y}{d x}=0 \\ &=6 x y+3 x^{2} \frac{d y}{d x}-3 y^{2} \frac{d y}{d x}=0 \end{aligned}

\begin{aligned} &=6 x y+\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=0 \\ &=\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}}=\frac{-6 x y}{3\left(x^{2}-y^{2}\right)} \end{aligned}

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

When $m_{1}=\frac{\left(x^{2}-y^{2}\right)}{2 x y}$ and  $m_{2}=\frac{-2 x y}{x^{2}-y^{2}}$

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

$=\frac{\left(x^{2}-y^{2}\right)}{2 x y} \times \frac{-2 x y}{x^{2}-y^{2}}=-1$

Two curves  $x^{3}-3 x y^{2}=-2 \;\; \& \;\; 3 x^{2} y-y^{3}=2$ intersect orthogonally.