Find the center and foci of the ellipse. $\frac { ( x - 2 ) ^ { 2 } } { 45 } + \frac { ( y - 8 ) ^ { 2 } } { 81 } = 1$
A)center: $( - 2 , - 8 )$ ; foci: $( - 2 , - 14 ) , ( - 2 , - 2 )$
B)center: $( 2,8 )$ ; foci: $( 2,2 ) , ( 2,14 )$
C)center: $( 2,8 )$ ; foci: $( 14,8 ) , ( 8,8 )$
D)center: $( - 2 , - 8 )$ ; foci: $( - 8,8 ) , ( - 14,8 )$
E)center: $( 2,8 )$ ; foci: $( - 8 , - 8 ) , ( - 14 , - 8 )$

Question

Find the center and foci of the ellipse. $\frac { ( x - 2 ) ^ { 2 } } { 45 } + \frac { ( y - 8 ) ^ { 2 } } { 81 } = 1$ 
A)center: $( - 2 , - 8 )$ ; foci: $( - 2 , - 14 ) , ( - 2 , - 2 )$
B)center: $( 2,8 )$ ; foci: $( 2,2 ) , ( 2,14 )$
C)center: $( 2,8 )$ ; foci: $( 14,8 ) , ( 8,8 )$
D)center: $( - 2 , - 8 )$ ; foci: $( - 8,8 ) , ( - 14,8 )$
E)center: $( 2,8 )$ ; foci: $( - 8 , - 8 ) , ( - 14 , - 8 )$

Quizplus · Accepted Answer

The center of the ellipse is given by the coordinates inside the parentheses, which are $(2, 8)$. The foci are calculated using the formula $c = \sqrt{a^2 - b^2}$, where $a^2$ is the larger denominator and $b^2$ is the smaller one. Here, $a^2 = 81$ and $b^2 = 45$, so $c = \sqrt{81 - 45} = \sqrt{36} = 6$. Since the larger denominator is under the $y$ term, the foci are vertical from the center, at $(2, 8 \pm 6)$, which are $(2, 2)$ and $(2, 14)$.

Find the Center and Foci of the Ellipse (x−2)245+(y−8)281=1\frac { ( x - 2 ) ^ { 2 } } { 45 } + \frac { ( y - 8 ) ^ { 2 } } { 81 } = 145(x−2)2​+81(y−8)2​=1

Find the Center and Foci of the Ellipse $\frac { ( x - 2 ) ^ { 2 } } { 45 } + \frac { ( y - 8 ) ^ { 2 } } { 81 } = 1$