The normals at three points $P, Q, R$ of the parabola $y^2 = 4ax$ meet in $(h, k).$ The centroid of triangle $PQR$ lies on :

$A(-1, 3)$ is a fixed point outside the parabola $y^2 = 4ax$ $(a > 0)$ and $P$ is a point moving on the parabola. The locus of point $Q$ which divides $AP$ in the ratio $3:2$ is a conic. Then the focus of that conic is

$AB$ is a chord of a parabola $y^2 = 4ax, (a > 0)$ with vertex $A$. $BC$ is drawn perpendicular to $AB$ meeting the axis at $C$. The projection of $BC$ on the axis of the parabola is

Let $P, Q,$ and $R$ be three co-normal points on the parabola $y^2 = 4ax$. Then the correct statement$(s)$ is/are:

The two parabolas $y^2 = 4x$ and $x^2 = 4y$ intersect at a point $P$,whose abscissa is not zero,such that

If the line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = -x - 2$ at $A$ and $B$,then $PA \cdot PB$ is equal to,where $P \equiv (\sqrt{3}, 0)$.