Question

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

Answer 1

We have $y^2=4xand{x}^2+y^2-6x+1=0$
Since both the curves touch each other at (1, 2) i.e., curves are passing through (1, 2).
$\therefore 2y.\frac{dy}{dx}=4and2x+2y\frac{dy}{dx}=6\Rightarrow \frac{dy}{dx}=\frac{4}{2y}and\frac{dy}{dx}=\frac{6-2x}{2y}\Rightarrow {\left(\frac{dy}{dx}\right)}_{1.2}=\frac{4}{4}=1and{\left(\frac{dy}{dx}\right)}_{1.2}=\frac{6-2.1}{2.2}=\frac{4}{4}=1\Rightarrow m_1=1and{m}_2=1$
Thus, we see that slope of both the curves are equal to each other i.e., $m_1=m_2=1$ at the point (1,2).
Hence both the curves touch each other.