Question

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

Answer 1

$x^2+2xy\sec \theta +y^2=0$

Applying quadratic formula by assuming $y$ as variable

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

here $a=1,b=2xsec\theta ,c=x^2$

$y=\frac{-2x\sec \theta \pm \sqrt{{4x}^2{\sec }^2\theta -4\left(1\right)\left(x^2\right)}}{2}y=\frac{-2x\sec \theta \pm 2x\sqrt{{\sec }^2\theta -1}}{2}y=-x\sec \theta \pm x\tan \theta y=-x\sec \theta +x\tan \theta ,y=-x\sec \theta -x\tan \theta .......\left(i\right)y=\frac{-x}{\cos \theta }\pm x\frac{\sin \theta }{\cos \theta }y=\frac{-x}{\cos \theta }+x\frac{\sin \theta }{\cos \theta },y=\frac{-x}{\cos \theta }-x\frac{\sin \theta }{\cos \theta }\Rightarrow (1-\sin \theta )x+\cos \theta y=0,(1+\sin \theta )x+\cos \theta y=0$

So the eqaution of lines are $(1-\sin \theta )x+\cos \theta y=0$ and $(1+\sin \theta )x+\cos \theta y=0$

From $\left(i\right)$

Slope of first line $m_1=-\sec \theta +\tan \theta$

Slope of second line $m_2=-\sec \theta -\tan \theta$

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

$\tan \alpha =\left|\frac{-\sec \theta +\tan \theta +\sec \theta +tan\theta }{1-(\tan \theta -\sec \theta )(\tan \theta +\sec \theta )}\right|=\left|\frac{2\tan \theta }{1-({\tan }^2\theta -{\sec }^2\theta )}\right|\tan \alpha =\left|\frac{2\tan \theta }{1-(-1)}\right|=\tan \theta \Rightarrow \alpha ={\tan }^{-1}\left(\tan \theta \right)=\theta$