The equations to the straight lines passing through the origin and making an angle $\alpha$ with the straight line $\mathrm{y+x=0}$  are given by $\mathrm{x^2+\mathrm{kxy} \sec 2 \alpha+y^2=0}$, where $\mathrm{k}$=Option: 1 4Option: 2 3Option: 3 2Option: 4 1

Let $\mathrm{y=m x}$ be the equation of the straight line passing through the origin and making an angle $\alpha$ with $\mathrm{y+x=0}$. Now, slope of line $\mathrm{y+x=0 ~is~ -1 }$.

\begin{aligned} &\mathrm{ \therefore \tan \alpha= \pm \frac{m+1}{1-m} \text { or } \tan ^2 \alpha=\frac{(m+1)^2}{(m-1)^2} \quad \text { [squaring] } }\\ &\mathrm{ \text { or } m^2\left(1-\tan ^2 \alpha\right)+2 m\left(1+\tan ^2 \alpha\right)+\left(1-\tan ^2 \alpha\right)=0} . \end{aligned}[squaring]

This is a quadratic equation in m, therefore there will be two values of m and hence two straight lines which make an angle $\alpha$ with $\mathrm{y+x=0}$. Let the two values of m be $\mathrm{m_1}$ and $\mathrm{m_2}$, then

$\mathrm{m_1+m_2=-\frac{2\left(1+\tan ^2 \alpha\right)}{1-\tan ^2 \alpha}=-2 \sec 2 \alpha \text { and } m_1 m_2=\frac{1-\tan ^2 \alpha}{1-\tan ^2 \alpha}=1 . }$

Now the joint equation of the straight line passing through the origin and making an angle $\alpha$ with $\mathrm{y+x=0}$ is

\begin{aligned} &\mathrm{ \left(y-m_1 x\right)\left(y-m_2 x\right)=0 }\\ &\mathrm{ \text { or } \quad y^2-x y\left(m_1+m_2\right)+m_1 m_2 x^2=0 \text { or } y^2+2 x y \sec 2 \alpha+x^2=0 .} \end{aligned}