Find the center, foci, and vertices of the ellipse.
-$$16 ( x + 1 ) ^ { 2 } + 9 ( y - 3 ) ^ { 2 } = 144$$
A) center at $( - 1,3 )$
foci at $( - 1,3 - \sqrt { 7 } ) , ( - 1,3 + \sqrt { 7 } )$
vertices at $( - 1,7 ) , ( - 1 , - 1 )$
B) center at $( 1,3 )$
foci at $( 1,3 - \sqrt { 7 } ) , ( 1,3 + \sqrt { 7 } )$
vertices at $( 1,7 ) , ( 1 , - 1 )$
C) center at $( 0,3 )$
foci at $( 0,3 - \sqrt { 7 } ) , ( 0,3 + \sqrt { 7 } )$
vertices at $( 0,7 ) , ( 0 , - 1 )$
D) center at $( 3 , - 1 )$
foci at $( 3 , - 1 - \sqrt { 7 } ) , ( 3 , - 1 + \sqrt { 7 } )$
vertices at $( 3,7 ) , ( 3 , - 1 )$

Question

Quizplus · Accepted Answer

The standard form of an ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $(h,k)$ is the center, $a$ is the semi-major axis, and $b$ is the semi-minor axis. For the given equation, dividing both sides by 144 gives $\frac{(x+1)^2}{9} + \frac{(y-3)^2}{16} = 1$, indicating the center is at $(-1,3)$. The larger denominator under $y$ suggests the major axis is vertical, so $a^2 = 16$ and $b^2 = 9$. The distance from the center to the foci along the major axis is $c$, where $c^2 = a^2 - b^2 = 16 - 9 = 7$, so $c = \sqrt{7}$. Thus, the foci are at $(-1, 3 \pm \sqrt{7})$, and the vertices, being $a$ units from the center along the major axis, are at $(-1, 3 \pm 4)$, which simplifies to $(-1,7)$ and $(-1,-1)$.

Find the Center, Foci, and Vertices of the Ellipse 16(x+1)2+9(y−3)2=14416 ( x + 1 ) ^ { 2 } + 9 ( y - 3 ) ^ { 2 } = 14416(x+1)2+9(y−3)2=144

Find the Center, Foci, and Vertices of the Ellipse $16 ( x + 1 ) ^ { 2 } + 9 ( y - 3 ) ^ { 2 } = 144$