If $=\mathrm{Xcos\theta }-\mathrm{Ysin} \mathrm{\theta }, \mathrm{y}=\mathrm{Xsin} \mathrm{\theta }+\mathrm{Ycos} \mathrm{\theta }$ and

 ${\mathrm{x}}^2+4\mathrm{xy}+{\mathrm{y}}^2={\mathrm{AX}}^2+{\mathrm{BY}}^2,$ for $0\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2},$ then $(\mathrm{A}, \mathrm{B})=$

${\mathrm{x}}^2+4\mathrm{xy}+{\mathrm{y}}^2=0$

transformed to ${\mathrm{Ax}}^2+{\mathrm{By}}^2$

eliminating xy term if a=b

$$\begin{gathered} 2\mathrm{\theta }=\frac{\mathrm{\pi }}{2} \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{4} \\ \mathrm{x}=\frac{\mathrm{x}-\mathrm{y}}{\sqrt{2}} \mathrm{y}=\frac{\mathrm{x}+\mathrm{y}}{\sqrt{2}} \\ {\mathrm{x}}^2+4\mathrm{xy}+{\mathrm{y}}^2={\left(\frac{\mathrm{x}-\mathrm{y}}{\sqrt{2}}\right)}^2+4\left(\frac{\mathrm{x}-\mathrm{y}}{\sqrt{2}}\right) \left(\frac{\mathrm{x}+\mathrm{y}}{\sqrt{2}}\right)+{\left(\frac{\mathrm{x}+\mathrm{y}}{\sqrt{2}}\right)}^2 \end{gathered}$$

$$\begin{gathered} =\frac{2\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)+4\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}{2} \\ ={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{x}}^2-2{\mathrm{y}}^2 \\ =3{\mathrm{x}}^2-{\mathrm{y}}^2 \\ (\mathrm{A}, \mathrm{B})=(3, -1) \end{gathered}$$