Question

If $2\theta$ is acute angle, then the acute angle between the two lines $x^2(cos\theta -sin\theta )+2hxycos\theta +y^2(cos\theta +sin\theta )=0$ is

• A. $2\theta$
• B. $\theta /2$
• C. $\theta /3$
• D. $\theta$

Answer 1

The correct option is D $\theta$

$x^2(\cos \theta -\sin \theta )+2xy\cos \theta +y^2(\cos \theta +\sin \theta )=0$

Let $y=mx$ be general equation of line that satisfies above equation.

$\therefore x^2(\cos \theta -\sin \theta )+2x\left(mx\right)\cos \theta +\left(mx{)}^2\right(\cos \theta +\sin \theta )=0\therefore m^2(\cos \theta +\sin \theta )+m(2\cos \theta )+(\cos \theta -\sin \theta )=0\to quadraticinm$

$\therefore$ sum of roots $(m_1+m_2)=\frac{-2\cos \theta }{\cos \theta +\sin \theta }$

$\therefore$ product of roots $\left(m_1{m}_2\right)=\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }$

$\therefore alpha$ beangle between the lines.

$\tan \left(\alpha \right)\frac{m_1-m_2}{1+m_1{m}_2}=\frac{\sqrt{(m_1+m_2{)}^2-4m_1{m}_2}}{1+m_1{m}_2}=\frac{\sqrt{\frac{4{\cos }^2\theta }{(\cos \theta +\sin \theta {)}^2-\frac{4(\cos \theta -\sin \theta )}{(\cos \theta +\sin \theta )}}}}{1+\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }}=\frac{\sqrt{4{\cos }^2\theta -4({\cos }^2\theta -{\sin }^2\theta )}}{\cos \theta +\sin \theta +\cos \theta -\sin \theta }=\frac{\sqrt{{\cos }^2\theta -{\cos }^2\theta +{\sin }^2\theta }}{\cos \theta }=\tan \theta \alpha =\theta$

Answer D