#### Need solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 7

Hence, prove, two curves  touch each otherat (1,2).

Hint –

Two curves touch each other if $m_{1}=m_{2}$ where $m_{1}$ and $m_{2}$  are the slopes of two curves.

Given –

\begin{aligned} &y^{2}=4 x---(1) \\ &x^{2}+y^{2}-6 x+1=0---(2) \end{aligned}

Consider First curve is \begin{aligned} &y^{2}=4 x \end{aligned}

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 y \frac{d y}{d x}=4 \\ &=\frac{d y}{d x}=\frac{4}{2 y} \\ &=\frac{d y}{d x}=\frac{2}{y} \\ &=m_{1}=\frac{d y}{d x}=\frac{2}{y}---(3) \end{aligned}

Second curve is $x^{2}+y^{2}=8$

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

At (1, 2) we get
\begin{aligned} &=\left(\frac{d y}{d x}\right)_{c_{1}}=\frac{2}{2}=1 \\ &=\left(\frac{d y}{d x}\right)_{c_{2}}=\frac{3-1}{2}=\frac{2}{2}=1 \end{aligned}

Clearly, $\left(\frac{d y}{d x}\right)_{c_{1}}=\left(\frac{d y}{d x}\right)_{c_{2}}$  at $(1,2)$

So, given two curves touch each other at (1,2).