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 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 $\pm a$, which gives $\pm 9$, so 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 $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 36 } = 1$