Find the center, foci, and vertices of the ellipse.
-$$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 } = 1$$
A) center at $( 0,0 )$
foci at $( 0 , - 3 )$ and $( 0,3 )$
vertices at $( 0 , - 9 ) , ( 0,9 )$
B) center at $( 0,0 )$
foci at $( 0 , - 6 \sqrt { 2 } )$ and $( 0,6 \sqrt { 2 } )$
vertices at $( 0 , - 9 ) , ( 0,9 )$
C) center at $( 0,0 )$
foci at $( - 6 \sqrt { 2 } , 0 )$ and $( 6 \sqrt { 2 } , 0 )$
vertices at $( - 9,0 ) , ( 9,0 )$
D) center at $( 0,0 )$
foci at $( - 9,0 )$ and $( 9,0 )$
vertices at $( - 81,0 ) , ( 81,0 )$

Question

Quizplus · Accepted Answer

The general form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes, respectively. For the given ellipse, $a^2 = 81$ and $b^2 = 9$, so $a = 9$ and $b = 3$. The center of the ellipse is at the origin $(0,0)$ because there are no shifts in the equation. The foci are calculated using $c^2 = a^2 - b^2$, which gives $c = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2}$. Since $a > b$, the ellipse is stretched along the x-axis, and the foci and vertices are along the x-axis. Therefore, the foci are at $(-6\sqrt{2}, 0)$ and $(6\sqrt{2}, 0)$, and the vertices are at $(-9, 0)$ and $(9, 0)$.

Find the Center, Foci, and Vertices of the Ellipse x281+y29=1\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 } = 181x2​+9y2​=1

Find the Center, Foci, and Vertices of the Ellipse $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 } = 1$