Question

Find what straight lines are represented by the following equation and determine the angles between them.
$x^2+2xy\cot \theta +y^2=0$

Answer 1

$\sqrt{\cos 2\theta }$

Applying quadratic formula by assuming $y$ as variable

Here $a=1,b=2x\cot \theta ,c=x^2$

$y=\frac{-2x\cot \theta \pm \sqrt{{4x}^2{\cot }^2\theta -4\left(1\right)\left(x^2\right)}}{2}y=\frac{-2x\cot \theta \pm 2x\sqrt{{\cot }^2\theta -1}}{2}y=\frac{-2x\cot \theta \pm 2x\sqrt{{cosec}^2\theta -1-1}}{2}y=\frac{-2x\cot \theta \pm 2x\sqrt{\frac{1}{{\sin }^2\theta }-2}}{2}=\frac{-2x\cot \theta \pm 2x\sqrt{\frac{1-2{\sin }^2\theta }{{\sin }^2\theta }}}{2}y=\frac{-2x\cot \theta \pm 2x\sqrt{\frac{\cos 2\theta }{{\sin }^2\theta }}}{2}=-x\cot \theta \pm x\frac{\sqrt{\cos 2\theta }}{\sin \theta }=-x\frac{\cos \theta }{\sin \theta }\pm x\frac{\sqrt{\cos 2\theta }}{\sin \theta }\Rightarrow \sin \theta y=-x\cos \theta \pm x\sqrt{\cos 2\theta }$

So the equation of lines are $ysin\theta +xcos\theta =x\sqrt{\cos 2\theta }$ and $ysin\theta +xcos\theta =-x\sqrt{\cos 2\theta }$

From $\left(i\right)$

Slope of first line $m_1=-\cot \theta +\frac{\sqrt{\cos 2\theta }}{\sin \theta }$

Slope of second line $m_2=-\cot \theta -\frac{\sqrt{\cos 2\theta }}{\sin \theta }$

Angle between two straight lines that is $\tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{\frac{-\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta }+\frac{\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta }}{1+(\frac{-\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta })(\frac{-\cos \theta }{\sin \theta }-\frac{\sqrt{\cos 2\theta }}{\sin \theta })}\right|\tan \theta =\left|\frac{\frac{-\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta }+\frac{\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta }}{1-(\frac{-\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta })(\frac{\cos \theta }{\sin \theta }+\frac{\sqrt{\cos 2\theta }}{\sin \theta })}\right|\tan \theta =\left|\frac{\frac{2\sqrt{\cos 2\theta }}{\sin \theta }}{\frac{{\sin }^2\theta -(\cos 2\theta -{\cos }^2\theta )}{{\sin }^2\theta }}\right|\tan \theta =\left|\frac{\frac{2\sqrt{\cos 2\theta }}{\sin \theta }}{\frac{{\sin }^2\theta -({\cos }^2\theta -{\sin }^2\theta -{\cos }^2\theta )}{{\sin }^2\theta }}\right|\tan \theta =\left|\frac{2\sin \theta \sqrt{\cos 2\theta }}{2{\sin }^2\theta }\right|=2cosec\theta \sqrt{\cos 2\theta }\Rightarrow \theta ={\tan }^{-1}\left(2cosec\theta \sqrt{\cos 2\theta }\right)$