The area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the extremities of its latus rectum is

If $S(a, b)$ is a fixed point and $P(\alpha, \beta)$ is a variable point such that $4[(x-a)^2+(y-b)^2]=(\alpha x+\beta y+7)^2$ represents a parabola,then the locus of $P(\alpha, \beta)$ is

$P$ is a point on the parabola $y^2 = 4ax$ $(a > 0)$ whose vertex is $A$. $PA$ is produced to meet the directrix in $D$ and $M$ is the foot of the perpendicular from $P$ on the directrix. If a circle is described on $MD$ as a diameter,then it intersects the $x$-axis at a point whose coordinates are:

If one of the vertices of an equilateral triangle inscribed in the parabola $y^2=12x$ coincides with the vertex of the parabola, then the area (in sq. units) of that triangle is (in $\sqrt{3}$)

The normal at the point $(2, 4)$ to the parabola $y^2 = 4ax$ meets the parabola again at another point. Find the coordinates of this point.

The length of the latus rectum of the parabola $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$ is