Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y =-\frac{1}{2}$. Then $S=\left\{x \in R : \tan ^{-1}\left(\sqrt{f(x)}+\sin ^{-1}(\sqrt{f(x)+1})\right)=\frac{\pi}{2}\right\}$:

If $x-2=t^2$ and $y=2t$ are the parametric equations of the parabola $y^2=a(x-b)$,then the value of $a+b$ equals

If $mx - y + c = 0$ is a normal at a point $P$ on the parabola $y^2 = 16x$ and the focal distance of $P$ is $40$ units,then $|c| =$

The Cartesian coordinates of the point on the parabola $y^2 = -16x$,whose parameter is $t = \frac{1}{2}$,are

Find the equation of the normal to the parabola $y^2 = 4x$ which is parallel to the line $y = 3x + 4$.

Assertion $(A)$: The curves $y^2 = 4x$ and $x^2 = -2y$ intersect at $(0,0)$ and $(2, -2)$ orthogonally.
Reason $(R)$: If the product of the slopes of the tangents drawn to two curves at their point of intersection is $-1$,then the curves are said to cut each other orthogonally. The correct option among the following is: