If $P$ is $(3, 1)$ and $Q$ is a point on the curve $y^2 = 8x$,then the locus of the mid-point of the line segment $PQ$ is

The focus of the parabola $x^2 = -16y$ is

Find the locus of the point of intersection of perpendicular tangents to the parabola $y^2 - 6y + 24x - 63 = 0$.

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:

If a normal chord of the parabola $y^2 = 4ax$ subtends a right angle at the vertex,its slope is-

Normals are drawn from the point $P(8,0)$ to the parabola $y^2=12x$. If $\theta$ is the acute angle between two non-horizontal normals among them,then $\tan \theta=$