Write the equation of the ellipse in standard form. Identify the center and vertices.
-$$4 x ^ { 2 } + 16 y ^ { 2 } - 40 x + 96 y + 180 = 0$$
A) $\frac { ( x - 5 ) ^ { 2 } } { 16 } + \frac { ( y + 3 ) ^ { 2 } } { 4 } = 1$
center: $( - 5,3 )$; vertices: $( - 9,3 ) , ( - 1,3 )$
B) $\frac { ( x - 5 ) ^ { 2 } } { 16 } + \frac { ( y + 3 ) ^ { 2 } } { 4 } = 1$
center: $( 5 , - 3 )$; vertices: $( 1 , - 3 ) , ( 9 , - 3 )$
C) $\frac { ( x - 5 ) ^ { 2 } } { 4 } + \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1$
center: $( - 5,3 )$; vertices: $( - 5,1 ) , ( - 5,5 )$
D) $\frac { ( x - 5 ) ^ { 2 } } { 4 } + \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1$
center: $( 5 , - 3 )$; vertices: $( 5 , - 5 ) , ( 5 , - 1 )$

Question

Quizplus · Accepted Answer

To convert the given equation to standard form, complete the square for both $x$ and $y$ terms. The equation rearranges to $\frac{(x-5)^2}{16} + \frac{(y+3)^2}{4} = 1$, indicating the center is at $(5, -3)$ and the major axis is horizontal, with vertices $(1, -3)$ and $(9, -3)$.

Write the Equation of the Ellipse in Standard Form 4x2+16y2−40x+96y+180=04 x ^ { 2 } + 16 y ^ { 2 } - 40 x + 96 y + 180 = 04x2+16y2−40x+96y+180=0

Write the Equation of the Ellipse in Standard Form $4 x ^ { 2 } + 16 y ^ { 2 } - 40 x + 96 y + 180 = 0$