Find the standard form of the equation of the ellipse with the given characteristics.
Vertices: $( 0,4 ) , ( 8,4 )$ ; minor axis of length 10.

A) $\frac { ( x + 4 ) ^ { 2 } } { 16 } + \frac { ( y + 4 ) ^ { 2 } } { 25 } = 1$
B) $\frac { ( x - 4 ) ^ { 2 } } { 16 } + \frac { ( y - 4 ) ^ { 2 } } { 25 } = 1$
C) $\frac { ( x + 4 ) ^ { 2 } } { 16 } + \frac { ( y - 4 ) ^ { 2 } } { 25 } = 1$
D) $\frac { ( x - 4 ) ^ { 2 } } { 16 } + \frac { ( y + 4 ) ^ { 2 } } { 25 } = 1$
E) $\frac { ( x - 4 ) ^ { 2 } } { 16 } - \frac { ( y - 4 ) ^ { 2 } } { 25 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics. 
Vertices: $( 0,4 ) , ( 8,4 )$ ; minor axis of length 10.

A) $\frac { ( x + 4 ) ^ { 2 } } { 16 } + \frac { ( y + 4 ) ^ { 2 } } { 25 } = 1$
B) $\frac { ( x - 4 ) ^ { 2 } } { 16 } + \frac { ( y - 4 ) ^ { 2 } } { 25 } = 1$
C) $\frac { ( x + 4 ) ^ { 2 } } { 16 } + \frac { ( y - 4 ) ^ { 2 } } { 25 } = 1$
D) $\frac { ( x - 4 ) ^ { 2 } } { 16 } + \frac { ( y + 4 ) ^ { 2 } } { 25 } = 1$
E) $\frac { ( x - 4 ) ^ { 2 } } { 16 } - \frac { ( y - 4 ) ^ { 2 } } { 25 } = 1$

Quizplus · Accepted Answer

The vertices given are along the x-axis, indicating the major axis is horizontal. The center of the ellipse is the midpoint of the vertices, which is $(4,4)$. The distance between the vertices is 8, so the major axis length is 8, and the semi-major axis length $a$ is half of that, $a=4$, giving $a^2=16$. The minor axis length is given as 10, so the semi-minor axis length $b$ is half of that, $b=5$, giving $b^2=25$. The standard form of the ellipse equation is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $(h,k)$ is the center. Thus, the correct equation is $\frac{(x-4)^2}{16} + \frac{(y-4)^2}{25} = 1$.

Find the Standard Form of the Equation of the Ellipse (0,4),(8,4)( 0,4 ) , ( 8,4 )(0,4),(8,4)

Find the Standard Form of the Equation of the Ellipse $( 0,4 ) , ( 8,4 )$