The ordinates of the vertices of a triangle inscribed in the parabola $y^2 = 4ax$ are $y_1, y_2, y_3$. Find the area of the triangle.

If the vertex of a parabola is $(4,3)$ and its directrix is $3x+2y-7=0$,then the equation of the latus rectum of the parabola is

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The graph of $y = f(x)$ represents a parabola whose focus has the coordinates:

If the line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = -x - 2$ at $A$ and $B$,then $PA \cdot PB$ is equal to,where $P \equiv (\sqrt{3}, 0)$.

At what points does the line $2x + y - 1 = 0$ intersect the parabola $y^2 = 4x$?

$PQ$ is a focal chord of the parabola $y^2 = 4x$ with focus $S$. If $P = (4, 4)$,then $SQ = $