Mathematics - Straight Lines Question with Solution | TestHub

Let $x^2+2hxy+y^2=0$ be given by $y=m_{1}x$ \& $y=m_{2}x.m_1+m_2=-2h$. Slope of $y+x=0$ is $-1.\tan \alpha =\left|\frac{m_1+1}{1-m_1}\right|,\tan \alpha =\left|\frac{m_2+1}{1-m_2}\right|.\tan \alpha =\frac{m_1+1}{1-m_1}\&\tan \alpha =-\frac{m_2+1}{1-m_2}$ (for +ve signs, in both gives the same value but $m_1\ne m_2)\Rightarrow m_1=\frac{\tan \alpha -1}{\tan \alpha +1},m_2=\frac{\tan \alpha +1}{\tan \alpha -1}.m_1+m_2=-2\sec 2\alpha \Rightarrow h=\sec 2\alpha$. $\cos 2\alpha =\frac{1}{h}.2{\cos }^2\alpha -1=\frac{1}{h}\Rightarrow \cos \alpha =\sqrt{\frac{1+h}{2h}}\Rightarrow \cot \alpha =\sqrt{\frac{h+1}{h-1}}$.