Find the center and vertices of the ellipse. $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 36 } = 1$
A)center: (8, 6) vertices: (-8, -6), (8, 6)
B)center: (0, 0) vertices: (0, -8), (0, 8)
C)center: (0, 0) vertices: (-8, 0), (8, 0)
D)center: (8, 0) vertices: (0, -6), (0, 6)
E)center: (0, 0) vertices: (-6, 0), (6, 0)

Question

Find the center and vertices of the ellipse. $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 36 } = 1$ 
A)center: (8, 6) vertices: (-8, -6), (8, 6)
B)center: (0, 0) vertices: (0, -8), (0, 8)
C)center: (0, 0) vertices: (-8, 0), (8, 0)
D)center: (8, 0) vertices: (0, -6), (0, 6)
E)center: (0, 0) vertices: (-6, 0), (6, 0)

Quizplus · Accepted Answer

The center of the ellipse given by the equation $\frac{x^2}{64} + \frac{y^2}{36} = 1$ is at the origin (0, 0) because there are no shifts in the x or y directions. The vertices along the x-axis are found by setting y=0 and solving for x, which gives $x = \pm \sqrt{64} = \pm 8$, hence the vertices are (-8, 0) and (8, 0).

Find the Center and Vertices of the Ellipse x264+y236=1\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 36 } = 164x2​+36y2​=1

Find the Center and Vertices of the Ellipse $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 36 } = 1$