Find the center and vertices of the ellipse. $x ^ { 2 } + 9 y ^ { 2 } + 16 x + 18 y + 64 = 0$
A)center: $( - 1 , - 8 )$ vertices: $( - 4 , - 8 ) , ( 9 , - 8 )$
B)center: $( 8,1 )$ vertices: $( 7,1 ) , ( 9,1 )$
C)center: $( - 8 , - 1 )$ vertices: $( - 9 , - 1 ) , ( - 7 , - 1 )$
D)center: $( - 8 , - 1 )$ vertices: $( - 11 , - 1 ) , ( - 5 , - 1 )$
E)center: $( 8,1 )$ vertices: $( 5,1 ) , ( - 11,1 )$

Question

Find the center and vertices of the ellipse.​ $x ^ { 2 } + 9 y ^ { 2 } + 16 x + 18 y + 64 = 0$ ​
A)center: $( - 1 , - 8 )$ vertices: $( - 4 , - 8 ) , ( 9 , - 8 )$
B)center: $( 8,1 )$ vertices: $( 7,1 ) , ( 9,1 )$
C)center: $( - 8 , - 1 )$ vertices: $( - 9 , - 1 ) , ( - 7 , - 1 )$
D)center: $( - 8 , - 1 )$ vertices: $( - 11 , - 1 ) , ( - 5 , - 1 )$
E)center: $( 8,1 )$ vertices: $( 5,1 ) , ( - 11,1 )$

Quizplus · Accepted Answer

To find the center and vertices of the ellipse given by the equation $x^2 + 9y^2 + 16x + 18y + 64 = 0$, we first complete the square for both $x$ and $y$ terms.Rearrange the equation:$$x^2 + 16x + 9y^2 + 18y = -64$$Complete the square for $x$ and $y$:$$x^2 + 16x + 64 + 9(y^2 + 2y + 1) = -64 + 64 + 9(1)$$$$x^2 + 16x + 64 + 9(y^2 + 2y + 1) = 0 + 9$$$$(x + 8)^2 + 9(y + 1)^2 = 9$$Divide by 9 to normalize the right side:$$\frac{(x + 8)^2}{9} + \frac{(y + 1)^2}{1} = 1$$From this, we can see that the center of the ellipse is at $(-8, -1)$.The major axis is along the $x$-direction because the denominator under the $x$-term is larger. The length of the major axis is $2\sqrt{9} = 6$, so the vertices are 3 units away from the center along the $x$-axis.Thus, the vertices are at $(-8 - 3, -1)$ and $(-8 + 3, -1)$, which simplifies to $(-11, -1)$ and $(-5, -1)$.Therefore, the correct answer is D) center: $(-8, -1)$, vertices: $(-11, -1), (-5, -1)$.

Find the Center and Vertices of the Ellipse x2+9y2+16x+18y+64=0x ^ { 2 } + 9 y ^ { 2 } + 16 x + 18 y + 64 = 0x2+9y2+16x+18y+64=0

Find the Center and Vertices of the Ellipse $x ^ { 2 } + 9 y ^ { 2 } + 16 x + 18 y + 64 = 0$