Graph Ellipses Not Centered at the Origin
-$$\frac { ( \mathrm { x } + 1 ) ^ { 2 } } { 36 } + \frac { ( \mathrm { y } + 3 ) ^ { 2 } } { 9 } = 1$$
A) foci at $( - 1 + 3 \sqrt { 3 } , - 3 )$ and $( - 1 - 3 \sqrt { 3 } , - 3 )$
B) foci at $( - 3 + 3 \sqrt { 3 } , - 1 )$ and $( - 3 - 3 \sqrt { 3 } , - 1 )$
C) foci at $( - 3 \sqrt { 3 } , - 3 )$ and $( 3 \sqrt { 3 } , - 3 )$
D) foci at $( - 1 + 3 \sqrt { 3 } , - 1 )$ and $( - 1 - 3 \sqrt { 3 } , - 1 )$

Question

Quizplus · Accepted Answer

The correct foci are calculated using the formula for an ellipse not centered at the origin. The given ellipse has its center at $(-1, -3)$, a major axis length of $2\sqrt{36} = 12$ (since $a^2 = 36$), and a minor axis length of $2\sqrt{9} = 6$ (since $b^2 = 9$). The distance of each focus from the center along the major axis is $c = \sqrt{a^2 - b^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}$. Since the larger denominator is under the $x$ term, the major axis is horizontal, placing the foci at $(-1 \pm 3\sqrt{3}, -3)$.

Graph Ellipses Not Centered at the Origin - (x+1)236+(y+3)29=1\frac { ( \mathrm { x } + 1 ) ^ { 2 } } { 36 } + \frac { ( \mathrm { y } + 3 ) ^ { 2 } } { 9 } = 136(x+1)2​+9(y+3)2​=1

Graph Ellipses Not Centered at the Origin
- $\frac { ( \mathrm { x } + 1 ) ^ { 2 } } { 36 } + \frac { ( \mathrm { y } + 3 ) ^ { 2 } } { 9 } = 1$