Find the center and vertices of the ellipse. $$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 36 } = 1$$
A)center: (9,6)vertices: (-9,-6), (9,6)
B)center: (0,0)vertices: (0,-9), (0,9)
C)center: (9,0)vertices: (0,-6), (0,6)
D)center: (0,0)vertices: (-6,0), (6,0)
E)center: (0,0)vertices: (-9,0), (9,0)

Question

Find the center and vertices of the ellipse.​ $$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 36 } = 1$$ ​
A)center: (9,6)vertices: (-9,-6), (9,6)
B)center: (0,0)vertices: (0,-9), (0,9)
C)center: (9,0)vertices: (0,-6), (0,6)
D)center: (0,0)vertices: (-6,0), (6,0)
E)center: (0,0)vertices: (-9,0), (9,0)

Quizplus · Accepted Answer

The center of the ellipse given by the equation $\frac{x^2}{81} + \frac{y^2}{36} = 1$ is at the origin (0,0), because the equation is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 81$ and $b^2 = 36$. The vertices along the x-axis are at $\pm\sqrt{81} = \pm9$, hence the vertices are (-9,0) and (9,0).

Find the Center and Vertices of the Ellipse x281+y236=1\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 36 } = 181x2​+36y2​=1

Find the Center and Vertices of the Ellipse $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 36 } = 1$