What is the length of the latus rectum of the parabola $x^2 = -y$?

The area (in sq. units) of an equilateral triangle inscribed in the parabola $y^{2}=8x$, with one of its vertices at the vertex of this parabola, is (in $\sqrt{3}$)

If $(2, k)$ is a point on the parabola passing through the points $(1, -3), (-1, 5), (0, 2)$ and having its axis parallel to the $Y$-axis,then $k$ is equal to

$A$ circle of radius $2$ unit passes through the vertex and the focus of the parabola $y^{2}=2x$ and touches the parabola $y=\left(x-\frac{1}{4}\right)^{2}+\alpha$,where $\alpha>0$. Then $(4\alpha-8)^{2}$ is equal to

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 11x$,then $2(m_1^2 + m_2^2) = $

Let the line $y - \sqrt{3}x + 3 = 0$ cut the parabola $2y^2 = 2x + 3$ at $A$ and $B$. If $P(\sqrt{3}, 0)$,then the value of $|PA - PB|$ is [where $PA$ denotes the distance between points $P$ and $A$].