Question

Transform the equation $2x^2+3\sqrt{3}xy+3y^2=2$ from axes inclined at ${30}^o$ to rectangular axes, the axis of $x$ remaining unchanged.

Answer 1

When we change cooridnate axes from one set at an angle $\omega$ to ${\omega }^{\prime }$ then

$x\sin \omega =x^{\prime }\sin (\omega -\theta )+y^{\prime }\sin (\omega -{\omega }^{\prime }-\theta )$

$y\sin \omega =x^{\prime }\sin \theta +y^{\prime }\cos ({\omega }^{\prime }+\theta )$

Here $\omega ={30}^{\circ },{\omega }^{\prime }={90}^{\circ }\theta =0^{\circ }$

$x\sin {30}^{\circ }=x^{\prime }\sin ({30}^{\circ }-0^{\circ })+y^{\prime }\sin ({30}^{\circ }-{90}^{\circ }-0^{\circ })\frac{x}{2}=-\frac{x^{\prime }}{2}-\frac{y^{\prime }\sqrt{3}}{2}x=-x^{\prime }-y^{\prime }\sqrt{3}x=-x^{\prime }-\sqrt{3}{y}^{\prime }y\sin {30}^{\circ }=x^{\prime }\sin 0^{\circ }+y^{\prime }\sin ({90}^{\circ }+0^{\circ })\frac{y}{2}=0+y^{\prime }y=2y^{\prime }$

Now given equation is

$2x^2-3\sqrt{3}xy+3y^2=22{(-x^{\prime }-\sqrt{3}{y}^{\prime })}^2+3\sqrt{3}(-x^{\prime }-\sqrt{3}{y}^{\prime })\left(2y^{\prime }\right)+3{\left(2y^{\prime }\right)}^2=22{x^{\prime }}^2+6{y^{\prime }}^2+4\sqrt{3}{x}^{\prime }{y}^{\prime }-6\sqrt{3}{x}^{\prime }{y}^{\prime }-18{y^{\prime }}^2+12{y^{\prime }}^2=22{x^{\prime }}^2-2\sqrt{3}{x}^{\prime }{y}^{\prime }-2=0$

${x^{\prime }}^2-\sqrt{3}{x}^{\prime }{y}^{\prime }-1=0$