The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x² such that the rectangle lies inside the parabola, is :

Let P(4, –4) and Q(9, 6) be two points on the parabola, y² = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of $$\Delta $$PXQ is maximum. Then this maximum area (in sq. units) is :

If the area of the triangle whose one vertex is at the vertex of the parabola, y² + 4(x – a²) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :

The length of the chord of the parabola x² $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$  is -