Question

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

• A. $H^2$
• B. $-H^2$
• C. $2H^2$
• D. $4H^2$

Answer 1

The correct option is D

$4H^2$

Explanation for correct option:

We know that the angle between pair of straight lines $A{x}^2+2Hxy+B{y}^2=0$ is $\tan \left(\theta \right)=\frac{2\sqrt{H^2-AB}}{A+B}$

Here, $\theta =\frac{\pi }{3}$ since the straight lines form an equilateral triangle.

$$\begin{gathered} \therefore \tan \left(\theta \right)=\frac{2\sqrt{H^2-AB}}{A+B} \\ \Rightarrow \tan \left(\frac{\pi }{3}\right)=\frac{2\sqrt{H^2-AB}}{A+B} \\ \Rightarrow \sqrt{3}=\frac{2\sqrt{H^2-AB}}{A+B} \\ \Rightarrow 3=\frac{4\left(H^2-AB\right)}{{\left(A+B\right)}^2} \\ \Rightarrow 3A^2+6AB+3B^2=4H^2-4AB \\ \Rightarrow 3A^2+10AB+3B^2=4H^2 \\ \Rightarrow 3A^2+9AB+AB+3B^2=4H^2 \\ \Rightarrow 3A\left(A+3B\right)+B\left(A+3B\right)=4H^2 \\ \Rightarrow \left(A+3B\right)\left(3A+B\right)=4H^2 \end{gathered}$$

Hence, option (D) $4H^2$, is the correct answer.