If the pair of straight lines given by Ax^2 + | vedclass

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(D) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $ax + by + c = 0$ if and only if the angle between the line

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$4H^2$

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(D) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $ax + by + c = 0$ if and only if the angle between the lines is $60^\circ$ and the lines are symmetric with respect to the line $ax + by + c = 0$.This condition is satisfied when the pair of lines is given by $(a^2 - 3b^2)x^2 + 8abxy + (b^2 - 3a^2)y^2 = 0$.Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we get:$A = a^2 - 3b^2$,$B = b^2 - 3a^2$,and $2H = 8ab$.Now,calculate $(A + 3B)(3A + B)$:$A + 3B = (a^2 - 3b^2) + 3(b^2 - 3a^2) = a^2 - 3b^2 + 3b^2 - 9a^2 = -8a^2$.$3A + B = 3(a^2 - 3b^2) + (b^2 - 3a^2) = 3a^2 - 9b^2 + b^2 - 3a^2 = -8b^2$.Therefore,$(A + 3B)(3A + B) = (-8a^2)(-8b^2) = 64a^2b^2$.Since $2H = 8ab$,we have $H = 4ab$,so $H^2 = 16a^2b^2$.Thus,$64a^2b^2 = 4(16a^2b^2) = 4H^2$.

If the pair of straight lines given by $Ax^2 + 2Hxy + By^2 = 0$ $(H^2 > AB)$ forms an equilateral triangle with the line $ax + by + c = 0$,then $(A + 3B)(3A + B)$ is equal to:

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