Find the center, foci, and vertices of the ellipse.
-$$\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$$
A) center at $ (0,0) $
foci at $ (0,8) $ and $ (4,0) $
vertices at $ (0,64),(16,0) $

B) center at $ (0,0) $
foci at $ (0,-8) $ and $ (0,8) $
vertices at $ (0,-64),(0,64) $

C)center at $ (0,0) $
foci at $ (-4 \sqrt{3}, 0) $ and $ (4 \sqrt{3}, 0) $
vertices at $ (-8,0),(8,0) $

D)center at $ (0,0) $
foci at $ (0,-4 \sqrt{3}) $ and $ (0,4 \sqrt{3}) $
vertices at $ (0,-8),(0,8) $

Question

Find the center, foci, and vertices of the ellipse.
-$$\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$$
A) center at $ (0,0) $
foci at $ (0,8) $ and $ (4,0) $
vertices at $ (0,64),(16,0) $

B) center at $ (0,0) $
foci at $ (0,-8) $ and $ (0,8) $
vertices at $ (0,-64),(0,64) $

C)center at $ (0,0) $
foci at $ (-4 \sqrt{3}, 0) $ and $ (4 \sqrt{3}, 0) $
vertices at $ (-8,0),(8,0) $

D)center at $ (0,0) $
foci at $ (0,-4 \sqrt{3}) $ and $ (0,4 \sqrt{3}) $
vertices at $ (0,-8),(0,8) $

Quizplus · Accepted Answer

The given ellipse equation $\frac{x^2}{16} + \frac{y^2}{64} = 1$ has its major axis along the y-axis, indicated by the larger denominator under $y^2$. The center of the ellipse is at the origin (0,0) because it is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ without any shifts. The vertices are located at a distance of $b = \sqrt{64} = 8$ units along the y-axis from the center, hence at (0,-8) and (0,8). The foci are found using the formula $c = \sqrt{b^2 - a^2}$, where $a^2 = 16$ and $b^2 = 64$, so $c = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}$. Therefore, the foci are at (0, -4\sqrt{3}) and (0, 4\sqrt{3}).

Find the Center, Foci, and Vertices of the Ellipse x216+y264=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 116x2​+64y2​=1

Find the Center, Foci, and Vertices of the Ellipse $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$