Question

If the lines represented by the equation $3y^2-x^2+2\sqrt{3}x-3=0$ are rotated about the point $(\sqrt{3},0)$ through an angle ${15}^0$, one in the clockwise direction and other in anti-clockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is

• A. $y^2-x^2+2\sqrt{3}x+3=0$
• B. $y^2-x^2+2\sqrt{3}x-3=0$
• C. $y^2-x^2-2\sqrt{3}x+3=0$
• D. $y^2-x^2+3=0$

Answer 1

The correct option is B $y^2-x^2+2\sqrt{3}x-3=0$

Given pair of equation of line:

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

Which can be written as,

$(\sqrt{3}y{)}^2=x^2-2\sqrt{3}x+3$
$\Rightarrow (\sqrt{3}y{)}^2=(x-\sqrt{3}{)}^2$
$\Rightarrow L_1:\sqrt{3}y=x-\sqrt{3}$

And $L_2:\sqrt{3}y=-(x-\sqrt{3})$

Here, $m_1=\frac{1}{\sqrt{3}},m_2=-\frac{1}{\sqrt{3}}$

Now the lines are rotated by ${15}^0$ about $(\sqrt{3},0)$

$\therefore$ New slopes are $m_1=1$ and $m_2=-1$

$L_1:y=1(x-\sqrt{3})$
$L_2:y=-1(x-\sqrt{3})$
Equation of pair of lines:

$y^2=(x-\sqrt{3}{)}^2$
$\Rightarrow y^2=x^2-2\sqrt{3x}+3$
$\Rightarrow y^2-x^2+2\sqrt{3}x-3=0$