Find the center and vertices of the ellipse.
$$x ^ { 2 } + 16 y ^ { 2 } - 2 x - 256 y + 1009 = 0$$
A) center: $( 8,1 ) \quad$ vertices: $( 4,1 ) , ( 12,1 )$
B) center: $( - 1 , - 8 ) \quad$ vertices: $( - 2 , - 8 ) , ( 0 , - 8 )$
C) center: $( 1,8 ) \quad$ vertices: $( 0,8 ) , ( 2,8 )$
D) center: $( 1,8 )$ vertices: $( - 3,8 ) , ( 5,8 )$
E) center: $( - 1 , - 8 ) \quad$ vertices: $( - 5 , - 8 ) , ( 3 , - 8 )$

Question

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. The given equation is:$$x^2 + 16y^2 - 2x - 256y + 1009 = 0$$Rearrange the equation by grouping $x$ and $y$ terms:$$x^2 - 2x + 16y^2 - 256y = -1009$$Complete the square for $x$ and $y$ terms:$$x^2 - 2x + 1 + 16(y^2 - 16y + 64) = -1009 + 1 + 16(64)$$$$x^2 - 2x + 1 + 16(y^2 - 16y + 64) = -1009 + 1 + 1024$$$$(x - 1)^2 + 16(y - 8)^2 = 16$$Divide the whole equation by 16 to get the standard form of the ellipse equation:$$\frac{(x - 1)^2}{16} + \frac{(y - 8)^2}{1} = 1$$From the standard form, we can identify the center of the ellipse as $(h, k) = (1, 8)$.The major axis is along the $x$-direction because the larger denominator is under the $x$ term. The length of the major axis is $2a$, where $a^2 = 16$, so $a = 4$. The vertices are found by adding and subtracting $a$ from the $x$-coordinate of the center, keeping the $y$-coordinate the same.Therefore, the vertices are at $(1 - 4, 8) = (-3, 8)$ and $(1 + 4, 8) = (5, 8)$.

Find the Center and Vertices of the Ellipse x2+16y2−2x−256y+1009=0x ^ { 2 } + 16 y ^ { 2 } - 2 x - 256 y + 1009 = 0x2+16y2−2x−256y+1009=0

Find the Center and Vertices of the Ellipse $x ^ { 2 } + 16 y ^ { 2 } - 2 x - 256 y + 1009 = 0$