#### Provide solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 3 sub question 1

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

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 –  $x^{2}=4y-------(1)$

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

The point of intersection of two curve (2,1).

First curve is $x^{2}=4y$

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=4 \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2} \\ &=m_{1}=\frac{d y}{d x}=\frac{x}{2}---(3) \end{aligned}
Second curve is $4y+x^{2}=8$

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

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

When m1=1 and  m2=-1

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

$= 1 \times -1 =-1$

Hence, two curves $x^{2}=4 y \;\;\;\& \;\;\; 4 y+x^{2}=8$ intersect orthogonally.