Problem:
Given the equation $10 x^2+x+3 y^2+6 y-9=0$, rewrite and classify the conic and state relevant properties.
Answer:
Classification: Ellipse
Equation: $10 \left(x+\frac{1}{20}\right)^2+3 (y+1)^2=\frac{481}{40}$
Foci: $\left(
\begin{array}{cc}
 -\frac{1}{20} & -1-\frac{\sqrt{\frac{3367}{3}}}{20} \\
 -\frac{1}{20} & \frac{1}{60} \left(\sqrt{10101}-60\right) \\
\end{array}
\right)$
Eccentricity: $\sqrt{\frac{7}{10}}$
Center: $\left\{-\frac{1}{20},\frac{1}{2} \left(-1-\frac{\sqrt{\frac{3367}{3}}}{20}+\frac{1}{60} \left(\sqrt{10101}-60\right)\right)\right\}$
Area Enclosed: $\frac{481 \pi }{40 \sqrt{30}}$
{'source': 'AMPS', 'file': 'amps/mathematica/algebra/conic_sections/30321.txt'}