If the parabola $y^2 = 4ax$ passes through the point $(1, -2)$,then the tangent at this point is

Find the equation of the parabola whose axis is parallel to the $y$-axis and which passes through the points $(0,4), (1,9)$ and $(4,5)$.

$A$ tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^{2} = 8x$,which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square,then $2a + b$ is equal to:

Normals are drawn from the point $P(8,0)$ to the parabola $y^2=12x$. If $\theta$ is the acute angle between two non-horizontal normals among them,then $\tan \theta=$

$A$ normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, -\alpha)$ to the parabola $x^2 = -4ay$,where $a > 0$. Let $L$ be the line passing through $(0, -\alpha)$ and parallel to the directrix of the parabola. Suppose that $L$ intersects the parabola at two points $A$ and $B$. Let $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$,then the value of $24a$ is. . . .

The point on the curve $y^2=2(x-3)$ at which the normal is parallel to the line $y-2x+1=0$ is