Question
The pair of lines $$\sqrt 3 {x^2} - 4xy + \sqrt 3 {y^2} = 0$$ are rotated about the origin by $$\frac{\pi }{6}$$ in the anticlockwise sense. The equation of the pair in the new position is :
A.
$$\sqrt 3 {x^2} - xy = 0$$
B.
$${x^2} - \sqrt 3 xy = 0$$
C.
$$xy - \sqrt 3 {y^2} = 0$$
D.
none of these
Answer :
$$\sqrt 3 {x^2} - xy = 0$$
Solution :
The lines are $$\left( {\sqrt 3 x - y} \right)\left( {x - \sqrt 3 y} \right) = 0$$
Their separate equations are $$y = \tan \,{60^ \circ }.x,\,y = \tan \,{30^ \circ }.x$$
After rotation, the separate equations are $$y = \tan \,{90^ \circ }.x,\,y = \tan \,{60^ \circ }.x$$
$$\therefore $$ the combined equation in the new position is $$x\left( {\sqrt 3 x - y} \right) = 0$$