Question

Transform the equation $x^2+xy+y^2=8$ from axes inclined at ${60}^o$ to axes bisecting the angles between the original axes.

Answer 1

To change one set of coordinate axes inclined at an angle $\omega$ to another system at ${\omega }^{\prime }$ and origin remain unchanged we substitute

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

Bisector of angle will make angle ${30}^{\circ }$ and ${120}^{\circ }$ with the axis

So here $\omega ={60}^{\circ },{\omega }^{\prime }={90}^{\circ }$ and $\theta ={30}^{\circ }$

$\Rightarrow x\sin {60}^{\circ }=x^{\prime }\sin ({60}^{\circ }-{30}^{\circ })+y^{\prime }\sin ({60}^{\circ }-{90}^{\circ }-{30}^{\circ })\Rightarrow \frac{\sqrt{3}}{2}x=\frac{1}{2}{x}^{\prime }-\frac{\sqrt{3}}{2}{y}^{\prime }\Rightarrow x=\frac{x^{\prime }}{\sqrt{3}}-y^{\prime }.......\left(i\right)$

Also, $y\sin {60}^{\circ }=x^{\prime }\sin {30}^{\circ }+y^{\prime }\sin ({90}^{\circ }+{30}^{\circ })$

$\Rightarrow \frac{\sqrt{3}}{2}y=\frac{1}{2}{x}^{\prime }+\frac{\sqrt{3}}{2}{y}^{\prime }\Rightarrow y=\frac{x^{\prime }}{\sqrt{3}}+y^{\prime }.......\left(ii\right)$

Given equation is

$x^2+xy+y^2=8$

Substituting $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow {(\frac{x^{\prime }}{\sqrt{3}}-y^{\prime })}^2+(\frac{x^{\prime }}{\sqrt{3}}-y^{\prime })(\frac{x^{\prime }}{\sqrt{3}}+y^{\prime })+{(\frac{x^{\prime }}{\sqrt{3}}+y^{\prime })}^2=8\Rightarrow \frac{{x^{\prime }}^2}{3}+{y^{\prime }}^2-\frac{2}{\sqrt{3}}{x}^{\prime }{y}^{\prime }+\frac{{x^{\prime }}^2}{3}+\frac{x^{\prime }{y}^{\prime }}{\sqrt{3}}-\frac{x^{\prime }{y}^{\prime }}{\sqrt{3}}-{y^{\prime }}^2+\frac{{x^{\prime }}^2}{3}+{y^{\prime }}^2+\frac{2}{\sqrt{3}}{x}^{\prime }{y}^{\prime }+\frac{{x^{\prime }}^2}{3}=8\Rightarrow {x^{\prime }}^2+{y^{\prime }}^2=8$