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 ${15}^{\circ }$, 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 $y^2-x^2+2\sqrt{3}x-3=0$

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 $y=\left(\tan {30}^{\circ }\right)x-1$ and $y=\left(\tan {150}^{\circ }\right)x+1$

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

$(y-0)=\tan {45}^{\circ }(x-\sqrt{3})$ and $(y-0)=\tan {135}^{\circ }(x-\sqrt{3})$

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

Their combined equation is
$(y-x+\sqrt{3})(y+x-\sqrt{3})=0$ or $y^2-x^2+2\sqrt{3}x-3=0$

Hence, option B.