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

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

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 –  $x^{2}=y-------(1)$

$x^{3}+6y=7-------(2)$

The point of intersection of two curve (1,1).

First curve is $x^{2}=y$

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=\frac{d y}{d x} \\ &=m_{1}=2 x---(3) \end{aligned}
Second curve is $x^{3}+6y=7$

Differentiating above with respect to x,
\begin{aligned} &=3 x^{2}+\frac{6 d y}{d x}=0 \\ &=\frac{d y}{d x}=\frac{-3 x^{2}}{6}=\frac{-x^{2}}{2} \\ &=m_{2}=\frac{-x^{2}}{2}---(4) \end{aligned}

Substituting (1, 1) for m1& m2 , we get,
\begin{aligned} &=m_{1}=2 x=2(1)=2 \\ &=m_{1}=2 \\ &=m_{2}=\frac{-x^{2}}{2}=\frac{(-1)^{2}}{2}=\frac{-1}{2} \\ &=m_{2}=\frac{-1}{2} \end{aligned}

When m1=2 and  m2=\begin{aligned}\frac{-1}{2} \end{aligned}

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

\begin{aligned}=2 \times \frac{-1}{2}=-1 \end{aligned}

Hence, two curves $x^{2}= y \;\;\;\& \;\;\; x^{3}+6y=7$ intersect orthogonally.