What is the equation of the tangent to the parabola $y^2 = 12x$ at the upper end of its latus rectum?

Suppose the parabola $(y-k)^2 = 4a(x-h)$ has vertex $A$ and passes through $O = (0,0)$ and $L = (0,2)$. Let $D$ be an end point of the latus rectum. Let the $Y$-axis intersect the axis of the parabola at $P$. Then,$\angle PDA$ is equal to

Three normals drawn from any point to the parabola $y^2 = 4ax$ cut the line $x = 2a$ in points whose ordinates are in arithmetical progression. Then the tangents of the angles which the normals make with the axis of the parabola are in:

What is the focus of the parabola $y^2 - x - 2y + 2 = 0$?

Tangents are drawn at points $(x_i, y_i); i = 1, 2, 3$ lying on the parabola $y = x^2$ to enclose a triangle of area $\Delta$. If $x_1, x_2, x_3$ form an increasing arithmetic progression,where $x_1 = -1$ and $y_3 = 9$,then $\Delta$ is ............. $sq. \, units$.

Let $A, B$ and $C$ be three points on the parabola $y^2=6x$ and let the line segment $AB$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left(\frac{AM \cdot BN}{CD}\right)^2$ is equal to...........