Find the equation of the parabola which passes through $(6,-2)$,has its vertex at the origin and its axis along the $y$-axis.

What is the slope of the normal to the parabola $x^2 + 4y = 0$ at the point $(2, -1)$?

The coordinates of the focus of the parabola $5x^2 = -12y$ are

If $PQ$ is a focal chord of the parabola $y^2=4x$ with focus $S$ and $P=(4,4)$,then $SQ=$

The equation of a tangent line to the parabola $y^2 = 8x$,which passes through the point $(1, 3)$,is:

Let the focal chord $PQ$ of the parabola $y^2=4x$ make an angle of $60^{\circ}$ with the positive $x$-axis,where $P$ lies in the first quadrant. If the circle,whose one diameter is $PS$,$S$ being the focus of the parabola,touches the $y$-axis at the point $(0, \alpha)$,then $5 \alpha^2$ is equal to :