Question

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

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

Answer 1

Answer: (D)

$4H^{2}$

Answer 2

We know that the pair of lines

$(a^{2} - 3b^{2})x^{2} + 8abxy + (b^{2} - 3a^{2})y^{2} = 0$with the line

.$ax + by + c = 0$form an equilateral triangle.

Hence comparing with $Ax^{2} + 2Hxy + By^{2} = 0$

then $A = a^{2} - 3b^{2},B = b^{2} - 3a^{2}$, $2H = 8ab$

Now $(A + 3B)(3A + B) = ( - 8a^{2})( - 8b^{2})$

⇒ $(8ab)^{2} = (2H)^{2} = 4H^{2}$