The point of intersection of the latus rectum and the axis of the parabola ${y^2} + 4x + 2y - 8 = 0$ is:

The shortest distance between the line $y - x = 1$ and the curve $x = y^2$ is

Three normals are drawn from the point $(3, 0)$ to the parabola $y^2 = 4x$,meeting the parabola at points $P, Q,$ and $R$. Match the following:

Column-$I$	Column-$II$
$(A)$ Circumradius of $\Delta PQR$	$(P)$ $5/2$
$(B)$ Area of $\Delta PQR$	$(Q)$ $(5/2, 0)$
$(C)$ Centroid of $\Delta PQR$	$(R)$ $(2/3, 0)$
$(D)$ Circumcenter of $\Delta PQR$	$(S)$ $2$

If the three normals drawn to the parabola $y^{2} = 2x$ pass through the point $(a, 0)$ where $a \neq 0$,then $a$ must be greater than:

Let $A(4, -4)$ and $B(9, 6)$ be points on the parabola $y^2 = 4x$. Let $C$ be a point chosen on the arc $AOB$ of the parabola,where $O$ is the origin,such that the area of $\Delta ACB$ is maximum. Then,the area (in sq. units) of $\Delta ACB$ is

Find the length of the subnormal of the parabola $y^2 = 16x$ at the point where the $x$-coordinate is $4$.