If the pair of straight lines given by $A x^{2} + 2 H x y + B y^{2} = 0 (H^{2} > A B)$ forms an equilateral triangle with line $a x + b y + c = 0$ , then $(A + 3 B) (3 A + B) =$

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Solution

$\begin{matrix} tan 60^{\circ} = \sqrt{3} = ∣ ∣ ∣ \frac{2 \sqrt{H^{2} - A B}}{A + B} ∣ ∣ ∣ 3 {(A + B)}^{2} = 4 (H^{2} - A B) 3 A^{2} + 3 B^{2} + 10 A B = 4 H^{2} (3 A + B) (A + 3 B) = 4 H^{2} . \end{matrix}$

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If the pair of straight lines given by Ax2+2Hxy+By2=0(H2>AB) forms an equilateral triangle with line ax+by+c=0, then (A+3B)(3A+B)=

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$\begin{matrix} tan 60^{\circ} = \sqrt{3} = ∣ ∣ ∣ \frac{2 \sqrt{H^{2} - A B}}{A + B} ∣ ∣ ∣ 3 {(A + B)}^{2} = 4 (H^{2} - A B) 3 A^{2} + 3 B^{2} + 10 A B = 4 H^{2} (3 A + B) (A + 3 B) = 4 H^{2} . \end{matrix}$