What is the area of the triangle inscribed in the parabola $y^2 = 4x$ whose vertices have $y$-coordinates $1, 2,$ and $4$?

The curve described parametrically by $x = t^2 - 2t + 2$ and $y = t^2 + 2t + 2$ represents:

Statement $(A)$: If the normal at the ends of the latus rectum of the parabola $y^2 = 4x$ meet the curve again at $P$ and $P'$,then $PP' = 12$ units.
Reason $(R)$: If the normal at $T_1$ to the parabola $y^2 = 4ax$ meets the parabola again at $T_2$,then $T_2 = -T_1 - \frac{2}{T_1}$.

The directrix of the parabola $2 y^2+25 x=0$ is $........$

The normal drawn at a point $(2, -4)$ on the parabola $y^2 = 8x$ cuts the same parabola again at $(\alpha, \beta)$. Then $\alpha + \beta =$

The locus of the mid-point of the line segment joining the focus and any point on the parabola $y^{2}=4ax$ is a parabola. Find the equation of its directrix.