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

If the equation of the normal to the curve $y = \frac{x-a}{(x+b)(x-2)}$ at the point $(1, -3)$ is $x - 4y = 13$,then the value of $a+b$ is equal to $.......$.

If the curve $y = ax^{2} + bx + c, x \in R,$ passes through the point $(1, 2)$ and the tangent line to this curve at the origin is $y = x$,then the possible values of $a, b, c$ are:

The point on the curve $y = 2 + \sqrt{4x + 1}$ at which the slope of the tangent is $\frac{2}{5}$ is:

Let $C$ be a curve given by $y(x)=1+\sqrt{4x-3}$,$x>\frac{3}{4}$. If $P$ is a point on $C$ such that the tangent at $P$ has a slope of $\frac{2}{3}$,then a point through which the normal at $P$ passes is:

For the curve $x = a(\cos \theta + \theta \sin \theta)$,$y = a(\sin \theta - \theta \cos \theta)$,which of the following is true for the normal at any point $\theta$?