If the normal at point $P(t)$ to the parabola $y^2 = 16x$ meets it again at point $Q(36, -24)$,then the maximum possible focal distance of point $P$ is-

The tangents to the parabola $y^2 = 4ax$ from an external point $P$ make angles $\theta_1$ and $\theta_2$ with the axis of the parabola,such that $\tan \theta_1 + \tan \theta_2 = b$,where $b$ is a constant. Then $P$ lies on

$A$ circle is drawn with its centre at the focus of the parabola $y^2 = 2px$ such that it touches the directrix of the parabola. Then a point of intersection of the circle and the parabola is

Let $P$ and $Q$ be distinct points on the parabola $y^2=2x$ such that a circle with $PQ$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $\Delta OPQ$ is $3\sqrt{2}$,then which of the following is (are) the coordinates of $P$?
$(A)$ $(4, 2\sqrt{2})$
$(B)$ $(9, 3\sqrt{2})$
$(C)$ $(\frac{1}{4}, \frac{1}{\sqrt{2}})$
$(D)$ $(1, \sqrt{2})$

Find the equation of the parabola with vertex $(-3, 0)$ and directrix $x + 5 = 0$.

If $L$ is the normal drawn to the parabola $y^2 = 8x$ at the point $t = \frac{1}{\sqrt{2}}$,then the foot of the perpendicular drawn from the focus of the parabola on to the normal $L$ is