The equation of the normal to the curve $y=x(2-x)$ at the point (2, 0)  is

Solution:

The equation of the curve is

$\begin{array}{l} y=x(2-x)\text{ or, }y=2x-x^2\\ \Rightarrow \frac{dy}{dx}=2-2x\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,0)}=2-2\times 2=-\left|2\right|\end{array}$

So, the equation of the normal at (2, 0) is

$y-0=-\frac{1}{-2}(x-2)$ or, $2y=x-2$