Rotate the Axes So That the New Equation Contains No $x^{2}+x y+y^{2}-3 y-6=0$

Rotate the axes so that the new equation contains no xy-term. Discuss the new equati
- $x^{2}+x y+y^{2}-3 y-6=0$

A)
$\begin{array}{l}\theta=45^{\circ} \\y^{\prime 2}=-18 x^{\prime} \\\text { parabola } \\\text { vertex at }(0,0) \\\text { focus at }\left(-\frac{9}{2}, 0\right) \end{array}$

B)
$$\theta=45^{\circ}$$
$$\frac{\left(x^{\prime}-\frac{\sqrt{2}}{2}\right) ^{2}}{5}+\frac{\left(y^{\prime}-\frac{3 \sqrt{2}}{2}\right) ^{2}}{15}=1$$
ellipse
center at $\left(\frac{\sqrt{2}}{2}, \frac{3 \sqrt{2}}{2}\right)$
major axis is $y^{\prime}$ -axis
vertices at $\left(\frac{\sqrt{2}}{2},-\frac{3 \sqrt{2}}{2}\right)$ and $\left(\frac{\sqrt{2}}{2}, \frac{9 \sqrt{2}}{3}\right)$

C)
$$\begin{array}{l}\theta=45^{\circ} \\\frac{x^{2}}{6}-\frac{y^{\prime 2}}{8}=1\end{array}$$
hyperbola
center at $(0,0)$
transverse axis is the $x^{\prime}$ -axis
vertices at $(\pm \sqrt{6}, 0)$

D)
$\begin{array}{l}\theta=45^{\circ} \\\frac{x^{2}}{3}+\frac{y^{\prime 2}}{4}=1 \\\text { ellipse } \\\text { center at }(0,0) \\\text { major axis is } y^{\prime} \text {-axis } \\\text { vertices at }(0, \pm 2) \end{array}$