Question

Find the value of $b$, if $\theta$ is the measure of the angle between the lines $3x^2+4xy+b{y}^2=0$ and $\tan \theta =\frac{1}{2}$.

Answer 1

Comparing it with $a{x}^2+2hxy+b{y}^2=0$

and

$m_1+m_2=-\frac{2h}{b}=\frac{-4}{b}{m}_1{m}_2=\frac{a}{b}=\frac{3}{b}$

$\tan \theta =\frac{1}{2}=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{(m_1+m_2{)}^2-4m_1{m}_2}}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{{(-\frac{2h}{b})}^2-4\frac{a}{b}}}{1+\frac{a}{b}}\right|=\left|\frac{\sqrt{{(-\frac{4}{b})}^2-4\frac{3}{b}}}{1+\frac{3}{b}}\right|\Rightarrow \frac{1}{2}=\left|\frac{\sqrt{{(-\frac{4}{b})}^2-4\frac{3}{b}}}{1+\frac{3}{b}}\right|\Rightarrow 1+\frac{3}{b}=2\sqrt{\frac{16}{b^2}-\frac{12b}{b^2}}\Rightarrow (1+\frac{3}{b}{)}^2=4(\frac{16}{b^2}-\frac{12b}{b^2})\Rightarrow b^2+9+6b=64-48b\Rightarrow b^2+54b-55=0\Rightarrow b=1,b=-55$