Question

Helpp!! ? Find the center, vertices, and foci of the ellipse with equation 3x^2 + 6y^2 = 18

Answer 1

Divide by 18,
\[\frac{x^2}{6}+\frac{y^2}{3}=1\Rightarrow\left(\frac{x}{\sqrt{6}}\right)^2+\left(\frac{y}{\sqrt{3}}\right)^2=1\]
Then the center is (0,0).
For the principal x-axis, the vertices are \[(-\sqrt{6},0), \ (\sqrt{6},0)\]
For the secondary y-axis, the vertices are \[(0,-\sqrt{3}), \ (0,\sqrt{3})\]
For the foci, solve,
\[c=\pm\sqrt{b^2-a^2}=\pm\sqrt{6-3}=\pm\sqrt{3}\]
So, the foci are
\[(-\sqrt{3},0), \ (\sqrt{3},0)\]