Find the center and foci of the ellipse.
$$\frac { ( x - 9 ) ^ { 2 } } { 32 } + \frac { ( y + 5 ) ^ { 2 } } { 36 } = 1$$
A) center: $( - 9,5 ) \quad$ foci: $( - 9,3 ) , ( - 9,7 )$
B) center: $( 9 , - 5 ) \quad$ foci: $( 9 , - 7 ) , ( 9 , - 3 )$
C) center: $( 9 , - 5 ) \quad$ foci: $( 7 , - 5 ) , ( 11 , - 5 )$
D) center: $( - 9,5 ) \quad$ foci: $( - 11 , - 5 ) , ( - 7 , - 5 )$
E) center: $( 9 , - 5 ) \quad$ foci: $( - 11,5 ) , ( - 7,5 )$

Question

Quizplus · Accepted Answer

The center of the ellipse is given by the point $(h, k)$ in the general equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Here, $h = 9$ and $k = -5$, so the center is $(9, -5)$. The foci are located along the major axis, which is vertical since $b^2 = 36$ is greater than $a^2 = 32$. The distance from the center to each focus is $c$, where $c^2 = b^2 - a^2$. Thus, $c^2 = 36 - 32 = 4$, so $c = 2$. The foci are therefore $(9, -5 \pm 2)$, or $(9, -7)$ and $(9, -3)$.

Find the Center and Foci of the Ellipse (x−9)232+(y+5)236=1\frac { ( x - 9 ) ^ { 2 } } { 32 } + \frac { ( y + 5 ) ^ { 2 } } { 36 } = 132(x−9)2​+36(y+5)2​=1

Find the Center and Foci of the Ellipse $\frac { ( x - 9 ) ^ { 2 } } { 32 } + \frac { ( y + 5 ) ^ { 2 } } { 36 } = 1$