If the pair of straight lines given by $A x^2+2 H x y+B y^2=0\left(H^2>A B\right)$ forms an equilateral…
If the pair of straight lines given by $A x^2+2 H x y+B y^2=0\left(H^2>A B\right)$ forms an equilateral triangle with line $a x+b y+c=0$, then $(A+3 B)(3 A+B)$ is equal to :
$H^2$
$-H^2$
$2 H^2$
$4 H^2$
Solution
Given that,
$A x^2+2 H x y+B y^2=0$...(i)
and $a x+b y+c=0$ ...(ii)
Since, triangle is equilateral, then angle between the two lines is $60^{\circ}$.
Angle between pair of lines is given by
$\cos 60^{\circ}=\frac{A+B}{\sqrt{(A-B)^2+4 H^2}}$
$\Rightarrow \quad \frac{A+B}{\sqrt{(A-B)^2+4 H^2}}=\frac{1}{2}$
$\Rightarrow \quad(A-B)^2+4 H^2=4(A+B)^2$
$\Rightarrow \quad 4\left(A^2+B^2+2 A B\right)-\left(A^2+B^2-2 A B\right)=4 H^2$
$\Rightarrow \quad 3\left(A^2+B^2\right)+10 A B=4 H^2$
$\Rightarrow \quad 3 A^2+10 A B+3 B^2=4 H^2$
$\therefore \quad(3 A+B)(A+3 B)=4 H^2$