The area of the trapezium whose vertices lie on the parabola `y^2 = 4x` and its diagonals pass through (1,0) and having length `25/4` units each is

The center of the circle which cuts parabola orthogonally the parabola y^2=4x at (1,2) passes through

The vertex of the parabola x^(2)-6x+4y+1=0 is

The area included between th parabolas y^(2)=4x and x^(2)=4y is

The area (in square unit ) bounded by the parabolas y^(2) = 4ax and x^(2) = 4ay is -

The area of the region bounded by the parabola y=x^(2)-4x+5 and the straight line y=x+1 is-

The parabola y^2=4ax passes through the point (2,-6). Find the length of the latus rectum.

The length of the latus rectum of the parabola x^2 −4x−8y+12=0 is

The fouce of the parabola x^(2)-6x+4y+1=0 is

A square has one vertex at the vertex of the parabola y^2=4a x and the diagonal through the vertex lies along the axis of the parabola. If the ends of the other diagonal lie on the parabola, the coordinates of the vertices of the square are (a) (4a ,4a) (b) (4a ,-4a) (c) (0,0) (d) (8a ,0)

If the area of the triangle whose one vertex is at the vertex of the parabola, y^(2) + 4 (x - a^(2)) = 0 and the other two vertices are the points of intersection of the parabola and Y-axis, is 250 sq units, then a value of 'a' is