If the area of the triangle whose one vertex is at the vertex of the parabola,${y^2} + 4(x - {a^2}) = 0$ and the other two vertices are the points of intersection of the parabola and $y$-axis,is $250 \text{ sq. units}$,then a value of $a$ is

Let $P$ be a point on the parabola $y^{2}=12x$ and $N$ be the foot of the perpendicular drawn from $P$ on the axis of the parabola. $A$ line is drawn through the mid-point $M$ of $PN$,parallel to the axis of the parabola,which meets the parabola at $Q$. If the $y$-intercept of the line $NQ$ is $\frac{4}{9}$,then:

If the normals at points $t_1$ and $t_2$ on the parabola $y^2 = 4ax$ intersect again on the parabola,then what is the value of $t_1t_2$?

If the tangent and normal at any point $P$ of a parabola meet the axis of the parabola in $T$ and $G$ respectively,then

Three points $O(0,0)$,$P(a, a^2)$,and $Q(-b, b^2)$ with $a > 0$ and $b > 0$ lie on the parabola $y = x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola,and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to:

The angle between the tangents drawn from the point $(1,4)$ to the parabola $y^2=4x$ is