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

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(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$. Since the triangle formed by these lines and $ax+by+c=0$ is equilateral,the angle between the li

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$4H^2$

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(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$. Since the triangle formed by these lines and $ax+by+c=0$ is equilateral,the angle between the lines represented by the pair must be $60^{\circ}$. The angle $	heta$ between the lines $Ax^2+2Hxy+By^2=0$ is given by $	an 	heta = \frac{2\sqrt{H^2-AB}}{|A+B|}$. For an equilateral triangle,$	heta = 60^{\circ}$,so $	an 60^{\circ} = \sqrt{3}$. Thus,$\sqrt{3} = \frac{2\sqrt{H^2-AB}}{|A+B|}$. Squaring both sides,we get $3 = \frac{4(H^2-AB)}{(A+B)^2}$. $3(A+B)^2 = 4H^2 - 4AB$. $3(A^2+B^2+2AB) = 4H^2 - 4AB$. $3A^2+3B^2+6AB = 4H^2 - 4AB$. $3A^2+10AB+3B^2 = 4H^2$. Factoring the left side,we get $(3A+B)(A+3B) = 4H^2$.

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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