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 is equal to $...$ sq. units.

If the line segment joining the vertex of the parabola $y^2=4ax$ and a point on the parabola makes an angle $\theta$ with the positive $X$-axis,then the length of that line segment is

If the equation of the parabola with vertex $V \left(\frac{3}{2}, 3\right)$ and the directrix $x + 2y = 0$ is $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$,then $\alpha + \beta + \gamma$ is equal to:

The tangents to the parabola $y^2 = 4ax$ from an external point $P$ make angles $\theta_1$ and $\theta_2$ with the axis of the parabola,such that $\tan \theta_1 + \tan \theta_2 = b$,where $b$ is a constant. Then $P$ lies on

The point on the parabola $2y = x^2$ which is nearest to the point $(0, 3)$ is

In the $XY$-plane,three distinct lines $l_1, l_2, l_3$ concur at a point $(\lambda, 0)$. Further,the lines $l_1, l_2, l_3$ are normals to the parabola $y^2=6x$ at the points $A=(x_1, y_1)$,$B=(x_2, y_2)$,and $C=(x_3, y_3)$ respectively. Then,we have: