Question

Find the joint equation of pair of line through the origin each of which makes an angle of ${60}^o$ with the $y$-axis.

Answer 1

Let $OA$ and $OB$ be the lines through the origin makin an angle of ${60}^o$ with the $Y-$ axis.
Then $OA$ and $OB$ make an angle of ${30}^o$ and ${150}^o$ with positive direction of $X-$axis.
$\therefore$ slope of $OA=\tan {30}^o=\frac{1}{\sqrt{3}}$
$\therefore$ equation of the line $OA$ is
$y=\frac{1}{\sqrt{3}}x$ i.e. $x-\sqrt{3y}=0$
Slope of $OB=\tan {150}^o=\tan ({180}^o-{30}^o)$
$=-\tan {30}^o=-\frac{1}{\sqrt{3}}$
$\therefore$ equation of the line $OB$ is
$y=-\frac{1}{\sqrt{3}}x$ i.e. $x+\sqrt{3y}=0$
$\therefore$ required combined equation is
$(x-\sqrt{3y})(x+\sqrt{3y})=0$
i.e. $x^2-3y^2=0$