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 clockwise direction and other in anti-clockwise direction, 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. None of these

Answer 1

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

The given equation of pair of straight lines can be rewritten as $(\sqrt{3}y-x+\sqrt{3})(\sqrt{3}y+x-\sqrt{3})=0$

Their separate equations are

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

or, $y=\frac{1}{\sqrt{3}}x-1$ and $y=-\frac{1}{\sqrt{3}}x+1$

or, $y=\left(\tan {30}^0\right)x-1$ and $y=\left(\tan {150}^0\right)x+1.$

After rotation through an angle of ${15}^0,$ the lines are

$(y-0)=\tan {45}^0(x-\sqrt{3})$

and, $(y-0)=\tan {135}^0(x-\sqrt{3})$

or, $y=x-\sqrt{3}$ and $y=-x+\sqrt{3}$

Their combined equation is

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

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