Find the center and vertices of the ellipse. $x ^ { 2 } + 16 y ^ { 2 } - 10 x - 32 y + 25 = 0$
A)center: $( 1,5 )$ ; vertices: $( - 3,5 ) , ( 5,5 )$
B)center: $( 5,1 )$ ; vertices: $( 1,1 ) , ( 9,1 )$
C)center: $( - 5 , - 1 )$ ; vertices: $( - 6 , - 1 ) , ( - 4 , - 1 )$
D)center: $( - 5 , - 1 )$ ; vertices: $( - 9 , - 1 ) , ( 1 , - 1 )$
E)center: $( 5,1 )$ ; vertices: $( 4,1 ) , ( 6,1 )$

Question

Find the center and vertices of the ellipse.  $x ^ { 2 } + 16 y ^ { 2 } - 10 x - 32 y + 25 = 0$ 
A)center: $( 1,5 )$ ; vertices: $( - 3,5 ) , ( 5,5 )$
B)center: $( 5,1 )$ ; vertices: $( 1,1 ) , ( 9,1 )$
C)center: $( - 5 , - 1 )$ ; vertices: $( - 6 , - 1 ) , ( - 4 , - 1 )$
D)center: $( - 5 , - 1 )$ ; vertices: $( - 9 , - 1 ) , ( 1 , - 1 )$
E)center: $( 5,1 )$ ; vertices: $( 4,1 ) , ( 6,1 )$

Quizplus · Accepted Answer

To find the center and vertices of the ellipse, we first complete the square for both $x$ and $y$ terms in the given equation $x^2 + 16y^2 - 10x - 32y + 25 = 0$.Rearrange the equation grouping $x$ and $y$ terms:$(x^2 - 10x) + 16(y^2 - 2y) = -25$Complete the square for $x$ and $y$ terms:$(x^2 - 10x + 25) + 16(y^2 - 2y + 1) = -25 + 25 + 16(1)$$(x - 5)^2 + 16(y - 1)^2 = 16$This is the standard form of an ellipse $(x - h)^2/a^2 + (y - k)^2/b^2 = 1$, where $(h, k)$ is the center of the ellipse. From the equation, the center is $(5, 1)$.For the vertices along the major axis, we look at the larger of the two denominators in the standard form. Since this ellipse is horizontally oriented (the larger coefficient is under the $x$ term), the vertices are found by adding and subtracting the square root of the corresponding denominator from the $x$-coordinate of the center. Here, $a^2 = 16$, so $a = 4$.Thus, the vertices are $(5 \pm 4, 1)$, which are $(1, 1)$ and $(9, 1)$.

Find the Center and Vertices of the Ellipse x2+16y2−10x−32y+25=0x ^ { 2 } + 16 y ^ { 2 } - 10 x - 32 y + 25 = 0x2+16y2−10x−32y+25=0

Find the Center and Vertices of the Ellipse $x ^ { 2 } + 16 y ^ { 2 } - 10 x - 32 y + 25 = 0$