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

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(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$. Let the angle between these lines be $2	heta$. Since the triangle formed with the line $ax+by+c

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$4H^2$

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(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$. Let the angle between these lines be $2	heta$. Since the triangle formed with the line $ax+by+c=0$ is equilateral,the angle between each pair of lines is $60^{\circ}$. The angle $	heta$ between the lines $y=m_1x$ and $y=m_2x$ is given by $	an 	heta = \left|\frac{2\sqrt{H^2-AB}}{A+B}ight|$. For an equilateral triangle,the angle between the lines is $60^{\circ}$,so $	heta = 30^{\circ}$. Thus,$	an 30^{\circ} = \frac{\sqrt{H^2-AB}}{|A+B|}$. Squaring both sides,we get $\frac{1}{3} = \frac{H^2-AB}{(A+B)^2}$. $(A+B)^2 = 3(H^2-AB)$. $A^2+B^2+2AB = 3H^2-3AB$. $A^2+B^2+5AB = 3H^2$. We need to evaluate $(A+3B)(3A+B) = 3A^2+AB+9AB+3B^2 = 3A^2+10AB+3B^2$. From the condition of the angle between the lines $Ax^2+2Hxy+By^2=0$ being $60^{\circ}$,we use the formula $\cos 60^{\circ} = \frac{|A+B|}{\sqrt{(A-B)^2+4H^2}}$. $\frac{1}{2} = \frac{|A+B|}{\sqrt{(A-B)^2+4H^2}} \implies (A-B)^2+4H^2 = 4(A+B)^2$. $A^2+B^2-2AB+4H^2 = 4(A^2+B^2+2AB)$. $4H^2 = 3A^2+3B^2+10AB$. Therefore,$(A+3B)(3A+B) = 3A^2+10AB+3B^2 = 4H^2$.

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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