MathematicsMedium127×since 2002Q3040If the pair of lines ax2+2(a+b)xy+by2=0a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0ax2+2(a+b)xy+by2=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 :A3a2−10ab+3b2=03{a^2} - 10ab + 3{b^2} = 03a2−10ab+3b2=0B3a2−2ab+3b2=03{a^2} - 2ab + 3{b^2} = 03a2−2ab+3b2=0C3a2+10ab+3b2=03{a^2} + 10ab + 3{b^2} = 03a2+10ab+3b2=0D3a2+2ab+3b2=03{a^2} + 2ab + 3{b^2} = 03a2+2ab+3b2=0Check answerSkip