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

Question

Quizplus · Accepted Answer

The center of the ellipse is given by the point $(h, k)$ in the general equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, which is $(3, -3)$ in this case. The lengths of the semi-major and semi-minor axes are $\sqrt{36} = 6$ and $\sqrt{9} = 3$, respectively. Since $a^2 = 36$ and $b^2 = 9$, and the ellipse is wider than it is tall, it is horizontally oriented. The foci are calculated using the formula $c = \sqrt{a^2 - b^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}$. Therefore, the foci are at $(3 \pm 3\sqrt{3}, -3)$. The vertices are at the ends of the major axis, which are $(3 \pm 6, -3)$, simplifying to $(-3, -3)$ and $(9, -3)$.

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

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