Rotate the axes so that the new equation contains no xy-term. Discuss the new equati
-$31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$

A)
$\begin{array}{l}
\theta=45^{\circ} \
x^{\prime 2}=-4 \sqrt{2} y^{\prime} \
\text { parabola } \
\text { vertex at }(0,0) \
\text { focus at }(0,-\sqrt{2}
\end{array}$

B)
$\begin{array}{l}
\theta=36.9^{\circ} \
\frac{x^{2}}{9}+\frac{y^{\prime 2}}{4}=1 \
\text { ellipse } \
\text { center at }(0,0) \
\text { major axis is } x^{\prime} \text {-axis } \
\text { vertices at }(\pm 3,0)
\end{array}$

C)
$\begin{array}{l}
\theta=30^{\circ} \
\frac{x^{\prime 2}}{4}+\frac{y^{\prime 2}}{9}=1 \
\text { ellipse } \
\text { center at }(0,0) \
\text { major axis is } y^{\prime} \text {-axi: } \
\text { vertices at }(0, \pm 3)
\end{array}$

D)
$\begin{array}{l}
\theta=45^{\circ} \
y^{\prime 2}=-4 \sqrt{2} x^{\prime} \
\text { parabola } \
\text { vertex at }(0,0) \
\text { focus at }(-\sqrt{2}, 0)
\end{array}$

Question

Rotate the axes so that the new equation contains no xy-term. Discuss the new equati
-$31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$

A)
$\begin{array}{l}
\theta=45^{\circ} \
x^{\prime 2}=-4 \sqrt{2} y^{\prime} \
\text { parabola } \
\text { vertex at }(0,0) \
\text { focus at }(0,-\sqrt{2}
\end{array}$

B)
$\begin{array}{l}
\theta=36.9^{\circ} \
\frac{x^{2}}{9}+\frac{y^{\prime 2}}{4}=1 \
\text { ellipse } \
\text { center at }(0,0) \
\text { major axis is } x^{\prime} \text {-axis } \
\text { vertices at }(\pm 3,0)
\end{array}$

C)
$\begin{array}{l}
\theta=30^{\circ} \
\frac{x^{\prime 2}}{4}+\frac{y^{\prime 2}}{9}=1 \
\text { ellipse } \
\text { center at }(0,0) \
\text { major axis is } y^{\prime} \text {-axi: } \
\text { vertices at }(0, \pm 3)
\end{array}$

D)
$\begin{array}{l}
\theta=45^{\circ} \
y^{\prime 2}=-4 \sqrt{2} x^{\prime} \
\text { parabola } \
\text { vertex at }(0,0) \
\text { focus at }(-\sqrt{2}, 0)
\end{array}$

Quizplus · Accepted Answer

The given equation is $31x^2 + 10\sqrt{3}xy + 21y^2 - 144 = 0$. To eliminate the $xy$ term, we use the formula for the angle of rotation $\theta$, which is given by $\tan(2\theta) = \frac{2B}{A-C}$, where $A$, $B$, and $C$ are the coefficients of $x^2$, $xy$, and $y^2$ respectively. Substituting the given values, we get $\tan(2\theta) = \frac{20\sqrt{3}}{10} = 2\sqrt{3}$, which gives $2\theta = 60^\circ$ or $\theta = 30^\circ$. After rotation, the equation becomes an ellipse because both squared terms are positive and have different coefficients, and there's no $xy$ term. The new equation is in the form $\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1$, indicating an ellipse centered at the origin. The major axis is along the $y'$ axis because $b^2 > a^2$, and the vertices are at $(0, \pm 3)$, confirming choice C.

Rotate the Axes So That the New Equation Contains No 31x2+103xy+21y2−144=031 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=031x2+103​xy+21y2−144=0

Rotate the Axes So That the New Equation Contains No $31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$