# Prove that the curves $y^2 = 4x$  and $x^2 + y^2 - 6x + 1 = 0$ touch each other at the point (1, 2).

### Answers (1)

Given: two curves $y^2 = 4x$  and $x^2 + y^2 - 6x + 1 = 0$

To prove: Two curves have the possibility of meeting at a point(1,2)

Explanation:

Now given $x^2 + y^2 - 6x + 1 = 0$

Differentiating the above equation with respect to x

$\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}+1\right)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(0)}{\mathrm{dx}}$
Now using the sum rule of differentiation

\begin{aligned} &\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(6 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(1)}{\mathrm{dx}}=0\\ &\Rightarrow 2 x+2 y \frac{d y}{d x}-6+0=0\\ &\Rightarrow 2 \mathrm{x}-6=-2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\\ &\Rightarrow 6-2 \mathrm{x}=2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\\ &\Rightarrow \frac{d y}{d x}=\frac{6-2 x}{2 y}\\ &\Rightarrow \frac{d y}{d x}=\frac{3-x}{y}\\ &\text { Finding the solution of above equation at point }(1,2) \text { , we get }\\ &\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,2)}=\frac{3-1}{2}=1=\mathrm{m}_{1} \ldots \ldots(i) \end{aligned}

Also given $y^2 = 4x$

Differentiating the value with respect to x, we get

$\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(4 \mathrm{x})}{\mathrm{dx}}$
Using a product rule of differentiation, we get
$\\\Rightarrow 2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=4 \\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}} \\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\mathrm{y}}=\mathrm{m}_{2}$
Finding the solution of the above equation at point (1,2), we get
$\Rightarrow\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}=1=m_{2} \ldots \ldots (ii)$

From equation (i) and (ii),

∴ $m_1 = m_2$

Therefore it is possible for both the curves to touch each other at points (1,2).

Hence proved

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