If the pair of lines ax^2 + 2(a + b)xy + by^2 | vedclass

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(B) The angle $	heta$ between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Here,$h =

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$3a^2 + 2ab + 3b^2 = 0$

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(B) The angle $	heta$ between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Here,$h = a + b$,so $	an 	heta = \left| \frac{2\sqrt{(a + b)^2 - ab}}{a + b} ight| = \left| \frac{2\sqrt{a^2 + ab + b^2}}{a + b} ight|$.The lines divide the circle into four sectors with angles $	heta$ and $\pi - 	heta$.Given that the area of one sector is thrice the area of another,the ratio of the angles is $1:3$.Thus,$	heta = \frac{\pi}{4}$ or $	heta = \frac{3\pi}{4}$ is not the correct approach; rather,the sectors are $	heta$ and $\pi - 	heta$. If one sector area is $3$ times another,then $\pi - 	heta = 3	heta$,which implies $4	heta = \pi$,so $	heta = \frac{\pi}{4}$.Therefore,$	an^2 	heta = 	an^2(\frac{\pi}{4}) = 1$.$\frac{4(a^2 + ab + b^2)}{(a + b)^2} = 1$.$4a^2 + 4ab + 4b^2 = a^2 + 2ab + b^2$.$3a^2 + 2ab + 3b^2 = 0$.

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 the sectors is thrice the area of another sector,then:

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