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

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(D) Let the angle between the two lines be $	heta$. The lines divide the circle into four sectors with angles $	heta$ and $\pi - 	heta$.Given that

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

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(D) Let the angle between the two lines be $	heta$. The lines divide the circle into four sectors with angles $	heta$ and $\pi - 	heta$.Given that the area of one sector is three times the area of the other,we have $\pi - 	heta = 3	heta$,which implies $4	heta = \pi$,so $	heta = \frac{\pi}{4} = 45^\circ$.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 45^\circ = \left| \frac{2\sqrt{(a + b)^2 - ab}}{a + b} ight|$.$1 = \frac{2\sqrt{a^2 + 2ab + b^2 - ab}}{a + b} = \frac{2\sqrt{a^2 + ab + b^2}}{a + b}$.Squaring both sides: $(a + b)^2 = 4(a^2 + ab + b^2)$.$a^2 + 2ab + b^2 = 4a^2 + 4ab + 4b^2$.$3a^2 + 2ab + 3b^2 = 0$.

If the pair of lines $ax^2 + 2(a + b)xy + by^2 = 0$ represent the diameters of a circle and divide the circle into four sectors such that the area of one sector is three times the area of the other,then:

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