Find the vertex of the parabola $y^2 + 6x - 2y + 13 = 0$.

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$ which is at minimum distance from the circle $x^2 + y^2 - 4x - 20y + 103 = 0$. Then $\alpha \beta$ is

The minimum distance between the parabola $y^2 = 8x$ and its image with respect to the line $x + y + 4 = 0$ is:

If the line $2x + y + k = 0$ is normal to the parabola $y^2 = -8x$,then the value of $k$ will be

Let $O$ be the vertex and $Q$ be any point on the parabola $x^2=8y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1:3$, then the locus of $P$ is

$A$ particle is moving in the $xy$-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the $x$-axis at $Q$. If the $y$-axis bisects the segment $PQ$,then $C$ is a parabola with