Prove that the curves

Solution:

Given:

Curves $y^{2}=4 x \ldots(1)$

$\& x^{2}+y^{2}-6 x+1=0$        ......(2)

$\therefore$ The point of intersection of two curves is $(1,2)$

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

Differentiating above w.r.t $\mathrm{x}$,

$\Rightarrow 2 y \cdot \frac{d y}{d x}=4$

$\Rightarrow y \cdot \frac{d y}{d x}=2$

$\Rightarrow m_{1}=\frac{2}{y} \ldots(3)$

Second curve is $x^{2}+y^{2}-6 x+1=0$

$\Rightarrow 2 x+2 y \cdot \frac{d y}{d x}-6-0=0$

$\Rightarrow x+y \cdot \frac{d y}{d x}-3=0$

$\Rightarrow y \cdot \frac{d y}{d x}=3-x$

$\Rightarrow \frac{d y}{d x}=\frac{3-x}{y} \ldots(4)$

At $(1,2)$, we have,

$m_{1}=\frac{2}{y}$

At $(1,2)$, we have,

$\Rightarrow m_{2}=\frac{3-x}{y}$

$\Rightarrow \frac{3-1}{2}$

$\Rightarrow m_{2}=1$

Clearly, $m_{1}=m_{2}=1$ at $(1,2)$

So, given curve touch each other at $(1,2)$