JEE AdvancedJEE Advanced 2008Application of Derivatives
Solution:
The given curves, are $C_{1} : y^{2} =4x .....\left(1\right)$ and $ C_{2} : x^{2} +y^{2} -6x + 1 = 0 .....\left(2\right)$
Solving (1) and (2) we get
$ x^{2} +4x-6x +1 =0 \Rightarrow x = 1$ and $ \Rightarrow y = 2$ or $-2 $
$\therefore $ Points of intersection of the two curves are $\left(1,2\right)$ and $ \left(1, -2\right)$
For $ C_{1} , \frac{dy}{dx} = \frac{2}{y} $
$ \therefore \left(\frac{dy}{dx}\right)_{\left(1,2\right)} = 1 =m_{1}$ and $\left(\frac{dy}{dx}\right)_{\left(1,-2\right)} = -1 = m_{1}' $
For $C_{2}, \frac{dy}{dx} = \frac{3-x}{y} \therefore \left(\frac{dy}{dx}\right)_{\left(1,2\right)} = 1 = m_{2}$
and $ \left(\frac{dy}{dx}\right)_{\left(1,-2\right)} = -1 =m_{2}' $
$\because m_{1} = m_{2} $ and $m_{1}' = m_{2}'$
$ \therefore C_{1} $ and $C_{2} $ touch each other at two points.