If 0 is the vertex and L,L' are the extremities of the latusrectum of the parabola `y^(2)=4ax` then the area of the triangle OLL' is

Find the area of the parabola y^(2)=4ax and the latusrectum.

Find the measure of the angle subtended by the latus rectum of the parabola y^(2) = 4ax at the vertex of the parabola .

The normals at the ends of the latusrectum of the parabola y^(2)=4ax" are (a, 2a) and (a, -2a)" .

Let A be the vertex and L the length of the latus rectum of the parabola,y^(2)-2y-4x-7=0 The equation of the parabola with A as vertex 2L the length of the latus rectum and the axis at right angles to that of the given curve is:

An isosceles triangle is inscribed in the parabola y^2 = 4ax with its base as the line joining the vertex and positive end of the latus rectum of the parabola. If (at^2, 2at) is the vertex of the triangle then

Let V be the vertex and L be the latusrectum of the parabola x^2=2y+4x-4 . Then the equation of the parabola whose vertex is at V. Latusrectum L//2 and axis s perpendicular to the axis of the given parabola.

If L and L' are ends of latus-rectum of parabola y^(2) = 4ax whose vertex is A , them m angle LAL ' =