The vertex of the parabola $9x^2 - 6x + 36y + 9 = 0$ is

Find the area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the ends of its latus rectum. (in $\text{ unit}^{2}$)

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 the tangent and normal at any point $P(at^2, 2at)$, $(a > 0)$, on the parabola $y^2 = 4ax$ meet the axis of the parabola at $T$ and $G$ respectively. Then the radius of the circle through $P, T$ and $G$ is

If the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again at $Q$,then the coordinates of $Q$ are

Suppose a parabola passes through $(0,4), (1,9)$ and $(4,5)$ and has its axis parallel to the $y$-axis. Then the equation of the parabola is