Find the center, foci, and vertices of the ellipse.
-$$4 x ^ { 2 } + 64 y ^ { 2 } = 256$$
A) center at $( 0,0 )$
foci at $( 0 , - 2 )$ and $( 0,2 )$
vertices at $( 0 , - 4 ) , ( 0,4 )$

B) center at $( 0,0 )$
foci at $( 0 , - 2 \sqrt { 15 } )$ and $( 0,2 \sqrt { 15 } )$
vertices at $( 0 , - 8 ) , ( 0,8 )$

C) center at $( 0,0 )$
foci at $( - 2 \sqrt { 15 } , 0 )$ and $( 2 \sqrt { 15 } , 0 )$
vertices at $( - 8,0 ) , ( 8,0 )$

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

Question

Find the center, foci, and vertices of the ellipse.
-$$4 x ^ { 2 } + 64 y ^ { 2 } = 256$$
A) center at $( 0,0 )$
foci at $( 0 , - 2 )$ and $( 0,2 )$
vertices at $( 0 , - 4 ) , ( 0,4 )$

B) center at $( 0,0 )$
foci at $( 0 , - 2 \sqrt { 15 } )$ and $( 0,2 \sqrt { 15 } )$
vertices at $( 0 , - 8 ) , ( 0,8 )$

C) center at $( 0,0 )$
foci at $( - 2 \sqrt { 15 } , 0 )$ and $( 2 \sqrt { 15 } , 0 )$
vertices at $( - 8,0 ) , ( 8,0 )$

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

Quizplus · Accepted Answer

The given equation of the ellipse is $4x^2 + 64y^2 = 256$. Dividing the entire equation by 256 gives $\frac{x^2}{64} + \frac{y^2}{4} = 1$, which is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 64$ and $b^2 = 4$. Since $a^2 > b^2$, the ellipse is oriented along the x-axis. The center is at (0,0), the vertices are at $(\pm a, 0) = (\pm 8, 0)$, and the foci are at $(\pm \sqrt{a^2 - b^2}, 0) = (\pm \sqrt{64 - 4}, 0) = (\pm \sqrt{60}, 0) = (\pm 2\sqrt{15}, 0)$.

Find the Center, Foci, and Vertices of the Ellipse 4x2+64y2=2564 x ^ { 2 } + 64 y ^ { 2 } = 2564x2+64y2=256

Find the Center, Foci, and Vertices of the Ellipse $4 x ^ { 2 } + 64 y ^ { 2 } = 256$