Question

lf the pair of lines $2x^2+3xy+y^2=0$ makes angles ${\theta }_1$ and ${\theta }_2$ with $x$-axis then $\tan ({\theta }_1-{\theta }_2)=$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{3}$

$2x^2+3xy+y^2=0$

Let $t=\frac{x}{y}$

$2t^2+3t+1=0$

$\Rightarrow t=\frac{-3\pm \sqrt{9-8}}{4}$

$\Rightarrow t=\frac{-3\pm 1}{4}$

$\Rightarrow t=-1,\frac{-1}{2}$

$\therefore y=-x$ and $y=-2x$

$\tan {\theta }_1=-1$ $\tan {\theta }_2=-2$

$\therefore tan({\theta }_1-{\theta }_2)=\frac{tan{\theta }_1-tan{\theta }_2}{1+tan{\theta }_{1}tan{\theta }_2}$

$\Rightarrow tan({\theta }_1-{\theta }_2)=\frac{-1+2}{1+2}=\frac{1}{3}$