If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$       lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then :

Solution :

As per question area of one sector $$= 3$$  area of another sector
$$ \Rightarrow $$  angle at centre by one sector $$ = 3 \times $$  angle at centre by another sector
Let one angle be $$\theta $$ then other $$ = 3\theta $$
Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$
$$\therefore $$  Angle between the diameters represented by combined equation $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0{\text{ is }}{45^ \circ }$$
$$\therefore $$  Using $$\tan \,\theta = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}}$$
we get, $$\tan \,{45^ \circ } = \frac{{2\sqrt {{{\left( {a + b} \right)}^2} - ab} }}{{a + b}}$$
$$\eqalign{ & \Rightarrow 1 = \frac{{2\sqrt {{a^2} + {b^2} + ab} }}{{a + b}} \cr & \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right) \cr & \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab \cr & \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0 \cr} $$