Let ABCD be a trapezium whose vertices lie on the parabola $\mathrm{y}^2=4 \mathrm{x}$. Let the sides AD and BC of the trapezium be parallel to $y$-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point $(1,0)$, then the area of $A B C D$ is

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

Let the shortest distance from $(a, 0), a>0$, to the parabola $y^2=4 x$ be 4 . Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is :

If the line $3 x-2 y+12=0$ intersects the parabola $4 y=3 x^2$ at the points $A$ and $B$, then at the vertex of the parabola, the line segment AB subtends an angle equal to