Question

The angle between the straight lines $x^2+4xy+y^2=0$ is?

• A. ${\tan }^{-1}\frac{1}{3}$
• B. ${\tan }^{-1}3$
• C. ${\tan }^{-1}\sqrt{\frac{3}{2}}$
• D. ${\tan }^{-1}\sqrt{3}$
• E. ${\tan }^{-1}1/\sqrt{3}$

Answer 1

Answer: (D)

${\tan }^{-1}\sqrt{3}$

Solution - $x^2+4xy+y^2=0$

General form: $a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0$

$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

In comparing we get

$a=1;b=1;h=2$ $\therefore \tan \theta =\left|\frac{2\sqrt{4-1}}{2}\right|=\sqrt{3}$ $\theta ={\tan }^{-1}\sqrt{3}$

Hence, $\left(D\right)$ is the correct option.