Find the center, foci, and vertices of the ellipse.
-$$16 ( x - 3 ) ^ { 2 } + 9 ( y - 1 ) ^ { 2 } = 144$$
A) center at $( 4,1 )$
foci at $( 4,1 - \sqrt { 7 } ) , ( 4,1 + \sqrt { 7 } )$
vertices at $( 4,5 ) , ( 4 , - 3 )$
B) center at $( 1,3 )$
foci at $( 1,3 - \sqrt { 7 } ) , ( 1,3 + \sqrt { 7 } )$
vertices at $( 1,5 ) , ( 1 , - 3 )$
C) center at $( - 3,1 )$
foci at $( - 3,1 - \sqrt { 7 } ) , ( - 3,1 + \sqrt { 7 } )$
vertices at $( - 3,5 ) , ( - 3 , - 3 )$
D) center at $( 3,1 )$
foci at $( 3,1 - \sqrt { 7 } ) , ( 3,1 + \sqrt { 7 } )$
vertices at $( 3,5 ) , ( 3 , - 3 )$

Question

Quizplus · Accepted Answer

The given equation is in the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $h=3$, $k=1$, $a^2=144/16=9$, and $b^2=144/9=16$. The center is at $(h,k)=(3,1)$. The major axis is vertical since $b^2 > a^2$, so the vertices and foci are above and below the center. The distance from the center to each vertex along the major axis is $\sqrt{b^2}=4$, and the distance from the center to each focus is $\sqrt{b^2-a^2}=\sqrt{7}$.

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

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