Find the center, foci, and vertices of the ellipse.
-$$\frac { ( x - 1 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$$
A) center at $ (1,-2) $
foci at $ (1+3 \sqrt{3}, 1),(1-3 \sqrt{3}, 1) $
vertices at $ (6,-2),(-6,-2) $

B)$\begin{array}{l}
\text { center at }(-2,1) \
\text { foci at }(-2+3 \sqrt{3}, 1),(-2-3 \sqrt{3}, 1) \
\text { vertices at }(-5,-2),(7,-2)
\end{array}$

C)center at $ (1,-2) $
foci at $ (-3 \sqrt{3},-2),(3 \sqrt{3},-2) $
vertices at $ (6,-2),(-6,-2) $

D) center at $ (1,-2) $
foci at $ (1+3 \sqrt{3},-2),(1-3 \sqrt{3},-2) $
vertices at $ (-5,-2),(7,-2) $

Question

Find the center, foci, and vertices of the ellipse.
-$$\frac { ( x - 1 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$$
A) center at $ (1,-2) $
foci at $ (1+3 \sqrt{3}, 1),(1-3 \sqrt{3}, 1) $
vertices at $ (6,-2),(-6,-2) $

B)$\begin{array}{l}
\text { center at }(-2,1) \
\text { foci at }(-2+3 \sqrt{3}, 1),(-2-3 \sqrt{3}, 1) \
\text { vertices at }(-5,-2),(7,-2)
\end{array}$

C)center at $ (1,-2) $
foci at $ (-3 \sqrt{3},-2),(3 \sqrt{3},-2) $
vertices at $ (6,-2),(-6,-2) $

D) center at $ (1,-2) $
foci at $ (1+3 \sqrt{3},-2),(1-3 \sqrt{3},-2) $
vertices at $ (-5,-2),(7,-2) $

Quizplus · Accepted Answer

The center of the ellipse is given by the point (h, k) in the general equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, which is (1, -2) in this case. The major axis is along the x-axis because $a^2 = 36$ is greater than $b^2 = 9$, so $a = 6$ and $b = 3$. The distance from the center to the foci is $c = \sqrt{a^2 - b^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}$. Therefore, the foci are at $(1 \pm 3\sqrt{3}, -2)$, and the vertices are at $(1 \pm 6, -2)$ or $(-5, -2)$ and $(7, -2)$.

Find the Center, Foci, and Vertices of the Ellipse (x−1)236+(y+2)29=1\frac { ( x - 1 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 136(x−1)2​+9(y+2)2​=1

Find the Center, Foci, and Vertices of the Ellipse $\frac { ( x - 1 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$