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

$A$ line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan^{-1}\left(\frac{3}{4}\right)$ with the $X$-axis. It intersects the parabola $y^{2}=4(x-3)$ at points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively. Then $|x_{1}-x_{2}|$ is equal to

Let $P$ be a point on the parabola $y^2 = 4ax$,where $a > 0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $PFQ$,where $F$ is the focus of the parabola,is $120$. If the slope $m$ of the normal and $a$ are both positive integers,then the pair $(a, m)$ is

$TP$ and $TQ$ are tangents of the parabola $y^2 = 8x$ at $P$ and $Q$ respectively. If the chord $PQ$ passes through the point $(-2, 3)$ and the locus of point $T$ is $y = mx + c$,then $(m + c)$ is equal to -

The tangents at the points $A(1, 3)$ and $B(1, -1)$ on the parabola $y^{2} - 2x - 2y = 1$ meet at the point $P$. Then the area (in unit$^{2}$) of the triangle $PAB$ is:

Statement $1$: $y = mx - \frac{1}{m}$ is always a tangent to the parabola $y^2 = -4x$ for all non-zero values of $m$.
Statement $2$: Every tangent to the parabola $y^2 = -4x$ will meet its axis at a point whose abscissa is non-negative.