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 1

Answer: (A)

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

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 $y=\frac{1}{\sqrt{3}}x$

$\Rightarrow y=\tan {60}^{0}x$ and $y=\tan {30}^{0}x$

After rotation, the separate equations are
$y=\tan {90}^{0}x$ and $y=\tan {60}^{0}x$

$\Rightarrow x=0$ and $y=\sqrt{3}x$

Therefore the combined equation in the new position is
$x(\sqrt{3}x-y)=0$ or $\sqrt{3}{x}^2-xy=0$