If (a + 3b)(3a + b) = 4h^2,then the angle bet | vedclass

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

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$60$

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(C) The angle $	heta$ between the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Given $(a + 3b)(3a + b) = 4h^2$,we expand the left side:$3a^2 + ab + 9ab + 3b^2 = 4h^2 \implies 3a^2 + 10ab + 3b^2 = 4h^2$.We know $4h^2 - 4ab = 3a^2 + 10ab + 3b^2 - 4ab = 3a^2 + 6ab + 3b^2 = 3(a + b)^2$.Therefore,$	an 	heta = \frac{\sqrt{4h^2 - 4ab}}{a + b} = \frac{\sqrt{3(a + b)^2}}{a + b} = \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,$	heta = 60^\circ$.

If $(a + 3b)(3a + b) = 4h^2$,then the angle between the lines represented by $ax^2 + 2hxy + by^2 = 0$ is.....$^o$

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