Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$. If $P$ also lies on the chord of the parabola $x^2 = 8y$ whose midpoint is $(1, 5/4)$,then $(\alpha - 28)(\beta - 8)$ is equal to:

$A$ point on the parabola whose focus and vertex are respectively at $\left(\frac{5}{4}, -2\right)$ and $(1, -2)$ is

Statement $-1:$ The line $x - 2y = 2$ meets the parabola $y^2 + 2x = 0$ only at the point $(-2, -2).$
Statement $-2:$ The line $y = mx - \frac{1}{2m}$ $(m \neq 0)$ is tangent to the parabola $y^2 = -2x$ at the point $\left( -\frac{1}{2m^2}, -\frac{1}{m} \right).$

The normal at a point on the parabola $y^2=4x$ passes through $(5,0)$. If there are two more normals to this parabola passing through $(5,0)$,then the equation of one of these normals is

The line $x - 1 = 0$ is the directrix of the parabola ${y^2} - kx + 8 = 0$. Then one of the values of $k$ is

The equation of the directrix of the parabola $y^2-x+4y+5=0$ is