Find the equation of the parabola that satisfies the following conditions: Focus $(0, -3)$,directrix $y = 3$.

The equation of the parabola with its vertex at the origin,axis on the $y$-axis and passing through the point $(6, -3)$ is

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y =-\frac{1}{2}$. Then $S=\left\{x \in R : \tan ^{-1}\left(\sqrt{f(x)}+\sin ^{-1}(\sqrt{f(x)+1})\right)=\frac{\pi}{2}\right\}$:

If $y = mx + c$ is the normal at a point on the parabola $y^2 = 8x$ whose focal distance is $8 \text{ units}$, then $|c|$ is equal to (in $\sqrt{3}$)

Let the directrix of the parabola $P : y^2 = 8x$, cut $x$-axis at the point $A$. Let $B(\alpha, \beta)$, $\alpha > 1$, be a point on $P$ such that the slope of $AB$ is $3/5$. If $BC$ is a focal chord of $P$, then six times the area of $\triangle ABC$ is :

$A = (-2, 0)$ and $P$ is a point on the parabola $y^2 = 8x$. If $Q$ bisects $\overline{AP}$ and the locus of $Q$ is a parabola,then its focus is