91Ó°ÊÓ

The given equation is$x^2+4xy+4y^2+5\sqrt{5}y+5=0$.

We need to determine the appropriate rotation formula so that new equation does not contain $\mathrm{xy}$term.

Here $A=1,B=4,C=4$

$cot2\mathrm{θ}=\frac{A-C}{B}=\frac{1-4}{4}=-\frac{3}{4}$

$$\begin{gathered} 2\mathrm{θ}=-53.12 \\ \mathrm{θ}=-26.56 \end{gathered}$$

$$\begin{gathered} x=x\text{'}\cos \left(-26.56\right)-y\text{'}\sin \left(-26.56\right) \\ x=0.8945x\text{'}+0.4471y\text{'} \\ and \\ y=x\text{'}\sin \left(-36.87\right)+y\text{'}\cos \left(-36.87\right) \\ y=-0.4471x\text{'}+0.8945y\text{'} \end{gathered}$$

$$\begin{gathered} x^2+4xy+4y^2+5\sqrt{5}y+5=0 \\ {\left(0.8945x\text{'}+0.4771y\text{'}\right)}^2+4\left(0.8945x\text{'}+0.4771y\text{'}\right)\left(-0.4771x\text{'}+0.8945y\text{'}\right)+4{\left(-0.4771x\text{'}+0.8945y\text{'}\right)}^2+5\sqrt{5}\left(-0.4771x\text{'}+0.8945y\text{'}\right)+5=0 \end{gathered}$$

$$\begin{gathered} 0.8x{\text{'}}^2+0.2y{\text{'}}^2+0.8x\text{'}y\text{'}+4\left(-0.4x{\text{'}}^2+0.8x\text{'}y\text{'}-0.2x\text{'}y\text{'}+0.4y{\text{'}}^2\right)+4\left(0.8y{\text{'}}^2+0.2x{\text{'}}^2-0.8x\text{'}y\text{'}\right)-2.3\sqrt{5}x\text{'}+4.4\sqrt{5}y\text{'}+5=0 \\ 0.8x{\text{'}}^2+0.2y{\text{'}}^2+0.8x\text{'}y\text{'}-1.6x{\text{'}}^2+3.2x\text{'}y\text{'}-0.8x\text{'}y\text{'}+1.6y{\text{'}}^2+3.2y{\text{'}}^2+0.8x{\text{'}}^2-3.2x\text{'}y\text{'}-2.3\sqrt{5}x\text{'}+4.4\sqrt{5}y\text{'}+5=0 \\ 0.8x{\text{'}}^2-1.6x{\text{'}}^2+0.8x{\text{'}}^2+0.2y{\text{'}}^2+1.6y{\text{'}}^2+3.2y{\text{'}}^2+0.8x\text{'}y\text{'}+3.2x\text{'}y\text{'}-0.8x\text{'}y\text{'}-3.2x\text{'}y\text{'}-2.3\sqrt{5}x\text{'}+4.4\sqrt{5}y\text{'}+5=0 \\ 0x{\text{'}}^2+5y{\text{'}}^2-2.3\sqrt{5}x\text{'}+4.4\sqrt{5}y\text{'}+5=0 \end{gathered}$$

This represent ellipse.