If the pair of straight line given by \(A x^2+2 H x y+B y^2=0\), where \(\left(H^2>A B\right)\), forms an…
If the pair of straight line given by \(A x^2+2 H x y+B y^2=0\), where \(\left(H^2>A B\right)\), forms an equilateral triangle with the line \(a x+b y+c=0\), then \((A+3 B)(3 A+B)=\)
\(4 H^2\)
\(2 H^2\)
\(-2 H^2\)
\(-4 H^2\)
Solution
According to the given information in the question the angle between lines represented by the equation \(A x^2+2 H x y+B y^2=0\) must be \(\frac{\pi}{3}\).
\(\begin{array}{ll}
\text {So, } & \tan \frac{\pi}{3}=\frac{2 \sqrt{H^2-A B}}{|A+B|} \\
\Rightarrow & 3(A+B)^2=4\left(H^2-A B\right) \\
\Rightarrow & 3 A^2+10 A B+3 B^2=4 H^2 \\
\Rightarrow & 3 A^2+9 A B+A B+3 B^2=4 H^2 \\
\Rightarrow & 3 A(A+3 B)+B(A+3 B)=4 H^2 \\
\Rightarrow & (A+3 B)(3 A+B)=4 H^2
\end{array}\)
Hence, option (a) is correct.