If the pair of lines ax^2 + 2(a+b)xy + by^2 = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of t - Mathematics MCQ AIEEE 2026 | ExamTest.live

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If the pair of lines a x^{2} + 2 (a + b) x y + 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 & Step-by-step Explanation

Let the angles of the sectors be θ and 3 θ .The two angles between a pair of intersecting lines sum to 18 0^{\circ} : θ + 3 θ = 18 0^{\circ} ⟹ 4 θ = 18 0^{\circ} ⟹ θ = 4 5^{\circ} .The angle between the pair of lines A x^{2} + 2 H x y + B y^{2} = 0 is given by: tan θ = \frac{2 H ^{2} - A B}{A + B} Here A = a, H = a + b, B = b : tan 4 5^{\circ} = \frac{2 ( a + b ) ^{2} - ab}{a + b} 1 = \frac{2 a ^{2} + 2 ab + b ^{2} - ab}{a + b} a + b = 2 a^{2} + ab + b^{2} Squaring both sides: (a + b)^{2} = 4 (a^{2} + ab + b^{2}) a^{2} + 2 ab + b^{2} = 4 a^{2} + 4 ab + 4 b^{2} 3 a^{2} + 2 ab + 3 b^{2} = 0 .

Try it yourself before checking the explanation above.

If the pair of lines a x^{2} + 2 (a + b) x y + 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:

3 a^{2} - 10 ab + 3 b^{2} = 0

3 a^{2} - 2 ab + 3 b^{2} = 0

3 a^{2} + 10 ab + 3 b^{2} = 0

3 a^{2} + 2 ab + 3 b^{2} = 0

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