Let $P$ be the point $(1, 0)$ and $Q$ be a point on the locus $y^2 = 8x$. The locus of the midpoint of $PQ$ is

If three points $P, Q, R$ on the parabola $y^2 = 4ax$ are such that their ordinates are in geometric progression,then the tangents at $P$ and $R$ intersect on:

Let $P$ be the point on the parabola ${y^2} = 8x$ which is at a minimum distance from the centre $C$ of the circle ${x^2} + {(y + 6)^2} = 1$. Then the equation of the circle,passing through $C$ and having its centre at $P$,is:

The points on the parabola $y^2 = 12x$ whose focal distance is $4$ are

Let the tangent to the parabola $S: y^{2}=2x$ at the point $P(2,2)$ meet the $x$-axis at $Q$ and the normal at $P$ meet the parabola $S$ at the point $R$. Then the area (in $sq. \ units$) of the triangle $PQR$ is equal to:

If $x^2 = 8ay$ is the transformed equation of $x^2 - 4y + 6x + 15 = 0$ when the origin is shifted to the point $(\alpha, \beta)$ by translation of axes,then $2\alpha + 8\beta^2 =$