The slope of the normal at the point $(at^2, 2at)$ of the parabola $y^2 = 4ax$ is

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:

$A$ beam is supported at its ends by supports which are $12 \, m$ apart. Since the load is concentrated at its centre,there is a deflection of $3 \, cm$ at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1 \, cm$?

For the parabola $y^2+6y-2x+5=0$,match the items in List-$I$ with the suitable item in List-$II$ given below:

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

The correct matching is:

The point on the curve $4y^2 - 4y + 2x - 1 = 0$ at which the tangent becomes parallel to the $Y$-axis is:

The condition that the parabolas $y^2 = 4ax$ and $y^2 = 4c(x - b)$ have a common normal other than the $x$-axis ($a, b, c$ being distinct positive real numbers) is