Find the center, foci, and vertices of the ellipse.
-$$\frac { ( x - 2 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$$
A) center at $( - 3,2 )$
foci at $( - 3 + 3 \sqrt { 3 } , 2 ) , ( - 3 - 3 \sqrt { 3 } , 2 )$
vertices at $( - 4 , - 3 ) , ( 8 , - 3 )$
B) center at $( 2 , - 3 )$
foci at $( 2 + 3 \sqrt { 3 } , 2 ) , ( 2 - 3 \sqrt { 3 } , 2 )$
vertices at $( 6 , - 3 ) , ( - 6 , - 3 )$
C) center at $( 2 , - 3 )$
foci at $( - 3 \sqrt { 3 } , - 3 ) , ( 3 \sqrt { 3 } , - 3 )$
vertices at $( 6 , - 3 ) , ( - 6 , - 3 )$
D) center at $( 2 , - 3 )$
foci at $( 2 + 3 \sqrt { 3 } , - 3 ) , ( 2 - 3 \sqrt { 3 } , - 3 )$
vertices at $( - 4 , - 3 ) , ( 8 , - 3 )$

Question

Quizplus · Accepted Answer

The center of the ellipse is given by the coordinates inside the parentheses, which are $(2, -3)$. The lengths of the semi-major and semi-minor axes are $\sqrt{36} = 6$ and $\sqrt{9} = 3$, respectively. The foci are calculated using the formula $c = \sqrt{a^2 - b^2}$, where $a = 6$ and $b = 3$, so $c = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}$. The foci are therefore at $(2 \pm 3\sqrt{3}, -3)$. The vertices are along the major axis, which is horizontal, and are 6 units away from the center, at $(2 \pm 6, -3)$ or $(-4, -3)$ and $(8, -3)$.

Find the Center, Foci, and Vertices of the Ellipse (x−2)236+(y+3)29=1\frac { ( x - 2 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 136(x−2)2​+9(y+3)2​=1

Find the Center, Foci, and Vertices of the Ellipse $\frac { ( x - 2 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$