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 because the denominator under $y^2$ is larger. The center of the ellipse is at the origin (0,0) because it is in the standard form. The length of the major axis is $2b = 2\sqrt{64} = 16$, so the vertices are at $(0, \pm8)$. The distance between the center and the foci is $c = \sqrt{b^2 - a^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}$, so the foci are at $(0, \pm4\sqrt{3})$.

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 $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$