The tangent to a given curve $y=f(x)$ is perpendicular to the $x$-axis,if

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ be twice the area $A_{2}$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does $NOT$ pass through the point.

Let $P(h, k)$ be a point on the curve $y=x^{2}+7x+2$ nearest to the line $y=3x-3$. Then the equation of the normal to the curve at $P$ is:

Find the angle of intersection of the curves $y^{2}=x$ and $x^{2}=y$.

Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$,where $x \in (0, \frac{\pi}{2})$. $A$ normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point:

The length of the normal at any point to the curve $y=c \cosh \left(\frac{x}{c}\right)$ is