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: $( 9 , - 5 )$ 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 coordinates inside the parentheses, which are $(9, -5)$. The foci are calculated using the formula $c = \sqrt{a^2 - b^2}$, where $a$ and $b$ are the semi-major and semi-minor axes, respectively. Here, $a^2 = 36$ and $b^2 = 32$, so $c = \sqrt{36 - 32} = 2$. Since the larger denominator is under $y$, the foci are vertical from the center, at $(9, -5 \pm 2)$, which simplifies to $(9, -7)$ 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 $\frac { ( x - 9 ) ^ { 2 } } { 32 } + \frac { ( y + 5 ) ^ { 2 } } { 36 } = 1$