The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 48y$ to the ends of its latus rectum 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 $\left( -\frac{1}{2m^2}, -\frac{1}{m} \right).$

$P$ 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:

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 $(4,5)$.

$A$ parabola has the origin $(0,0)$ as its focus and the line $x = 2$ as the directrix. Then the vertex of the parabola is at

Let $P$ be the point $(1, 0)$ and $Q$ be a point on the parabola $y^2 = 8x$. Find the locus of the midpoint of $PQ$.