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 :

A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is :

All x satisfying the inequality (cot^–1 x)²– 7(cot^–1 x) + 10 > 0, lie in the interval :

Let $$\sqrt 3 \widehat i + \widehat j,$$    $$\widehat i + \sqrt 3 \widehat j$$  and   $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \over {\sqrt 2 }}$$, then the sum of all possible values of $$\beta $$ is :