The equation of the latus rectum of the parabola represented by the equation ${y^2} + 2Ax + 2By + C = 0$ is

At what points does the line $2x + y - 1 = 0$ intersect the parabola $y^2 = 4x$?

Statement $- 1 :$ For all non-zero values of $m$,$y = mx - 1/m$ is always a tangent to the parabola $y^2 = -4x$.
Statement $- 2 :$ Every tangent to the parabola $y^2 = -4x$ touches its axis at a point whose $x$-coordinate is non-negative.

Normals at $P, Q, R$ are drawn to the parabola $y^2 = 4x$ which intersect at the point $(3, 0)$. Then,$\Delta PQR$ is:

If the line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = -x - 2$ at $A$ and $B$,then $PA \cdot PB$ is equal to,where $P \equiv (\sqrt{3}, 0)$.

Let $P$ be a point on the parabola $y^{2}=12x$ and $N$ be the foot of the perpendicular drawn from $P$ on the axis of the parabola. $A$ line is drawn through the mid-point $M$ of $PN$,parallel to the axis of the parabola,which meets the parabola at $Q$. If the $y$-intercept of the line $NQ$ is $\frac{4}{9}$,then: