Question

Prove that the equations to the straight lines passing through the origin which make an angle $\alpha$ with the straight line $y+x=0$ are given by the equation
$x^2+2xy\sec 2\alpha +y^2=0.$

Answer 1

The lines makes an angle $\alpha$ with the line $y+x=0$

Slope of given line that is $m_1$$=-1$

Let the slope of the other line be $m_2$

The angle between straight lines thatt is $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \alpha =\left|\frac{-1-m_2}{1-1.m_2}\right|=\left|\frac{m_2+1}{m_2-1}\right|\frac{m_2+1}{m_2-1}=\tan \alpha ,\frac{m_2+1}{m_2-1}=-\tan \alpha \Rightarrow m_2=\frac{\tan \alpha +1}{\tan \alpha -1},\frac{\tan \alpha -1}{\tan \alpha +1}$

Equation of straght line with given slope and a point is $y=mx+c$

Lines passes through origin $0=0.m+c$

$\Rightarrow c=0$

So the equation of lines are $y=\frac{\tan \alpha +1}{\tan \alpha -1}x$ and $y=\frac{\tan \alpha -1}{\tan \alpha +1}x$

$y=\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }x,y=\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }x$

Combined equation of straight lines

$(y-\frac{\sin \alpha +\cos \alpha }{sin\alpha -cos\alpha }x)(y-\frac{sin\alpha -cos\alpha }{\sin \alpha +\cos \alpha }x)=0y^2-xy\left(\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\right)-xy\left(\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }\right)+\left(\frac{\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha }\right)\left(\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\right)x^2=0y^2-xy\left(\frac{{(\sin \alpha -\cos \alpha )}^2+{(\sin \alpha +\cos \alpha )}^2}{\left(\sin \alpha +\cos \alpha \right)(\sin \alpha -\cos \alpha )}\right)+x^2=0y^2-xy\left(\frac{2}{{\sin }^2\alpha -{\cos }^2\alpha }\right)+x^2=0y^2-xy\left(\frac{2}{-\cos 2\alpha }\right)+x^2=0y^2+2xy\sec 2\alpha +x^2=0$

Hence proved