Question

The angle between the lines represented by the equation $ax^{2} + xy + by^{2} = 0$ will be $45^{o}$, if

• A. $a = 1,b = 6$
• B. $a = 1,b = - 6$
• C. $a = 6,b = 1$
• D. None of these

Answer 1

Answer: (B)

$a = 1,b = - 6$

Answer 2

$\tan 45{^\circ} = \frac{2\sqrt{\frac{1}{4} - ab}}{a + b}$

$\Rightarrow (a + b)^{2} = (1 - 4ab) \Rightarrow a^{2} + b^{2} + 6ab - 1 = 0$,

Which is obviously satisfied by $a = 1$and $b = - 6$.