# Question #b0b5a

Nov 30, 2017

$\sqrt{3} {x}^{2} - x y = 0.$

#### Explanation:

Let $S : \sqrt{3} {x}^{2} - 4 x y + \sqrt{3} {y}^{2} = 0.$

$\therefore \sqrt{3} {x}^{2} - 3 x y - x y + \sqrt{3} {y}^{2} = 0.$

$\therefore \sqrt{3} x \left(x - \sqrt{3} y\right) - y \left(x - \sqrt{3} y\right) = 0.$

$\therefore \left(\sqrt{3} x - y\right) \left(x - \sqrt{3} y\right) = 0.$

Thus, the individual lines ${l}_{1} \mathmr{and} {l}_{2} \text{ of } S$ are,

${l}_{1} : \sqrt{3} x - y = 0 , \mathmr{and} , {l}_{2} : x - \sqrt{3} y = 0.$

Observe that, $O \left(0 , 0\right) \in {l}_{1} : y = \sqrt{3} x = x \tan \left(\frac{\pi}{3}\right) ,$ and, is

making an $\angle \text{ of } \frac{\pi}{3}$ with the $+ v e$ direction of the $X -$ Axis.

So, if $S$ is rotated anti-clockwise by an $\angle \text{ of } \frac{\pi}{6}$ about $O$ , then, so

would be the effect on both ${l}_{1} \mathmr{and} {l}_{2.}$

This means that, after the said rotation, ${l}_{1}$ will now make an

$\angle \text{ of } \left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \frac{\pi}{2}$ with the $+ v e$ direction of the $X -$ Axis.

In other words, it will become the $Y -$Axis, or, $x = 0.$

Similarly, ${l}_{2}$ becomes, $y = x \tan \left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \sqrt{3} x .$

Hence, the combined eqn. of lines becomes

$x \left(\sqrt{3} x - y\right) = 0 , i . e . , \sqrt{3} {x}^{2} - x y = 0.$

Enjoy Maths.!