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. $x^2-\sqrt{3}xy=0$
• B. $xy-\sqrt{3}{y}^2=0$
• C. $\sqrt{3}{x}^2-xy=0$
• D. None of the above

Answer 1

Answer: (C)

The given equation of pair of straight lines can be rewritten as
$(\sqrt{3}x-y)(x-\sqrt{3}y)=0$.
Their separate equations are,
$y=\sqrt{3}x$ and $y=\frac{1}{\sqrt{3}}x$
$\Rightarrow y=\tan 60^{\circ }x$ and $y=\tan 30^{\circ }x$
After rotation, the separate equations are
$y=\tan 90^{\circ }x$ and $y=\tan 60^{\circ }x$
$\Rightarrow x=0$ and $y=\sqrt{3}\cdot x$
$\therefore$ Combined equation in the new position is
$x(\sqrt{3}x-y)=0$ or $\sqrt{3}{x}^2-xy=0$