Find the standard form of the equation of the ellipse with the given characteristics.
Vertices: (6,1), (6,17), minor axis of length 3

A) $\frac { ( x + 6 ) ^ { 2 } } { 9 } + \frac { ( y + 9 ) ^ { 2 } } { 64 } = 1$
B) $\frac { ( x - 6 ) ^ { 2 } } { 9 } + \frac { ( y - 9 ) ^ { 2 } } { 64 } = 1$
C) $\frac { ( x - 6 ) ^ { 2 } } { 64 } + \frac { ( y - 9 ) ^ { 2 } } { 9 } = 1$
D) $\frac { ( x - 9 ) ^ { 2 } } { 64 } + \frac { ( y - 6 ) ^ { 2 } } { 9 } = 1$
E) $\frac { ( x + 9 ) ^ { 2 } } { 64 } + \frac { ( y + 6 ) ^ { 2 } } { 9 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics. 
Vertices: (6,1), (6,17), minor axis of length 3

A) $\frac { ( x + 6 ) ^ { 2 } } { 9 } + \frac { ( y + 9 ) ^ { 2 } } { 64 } = 1$
B) $\frac { ( x - 6 ) ^ { 2 } } { 9 } + \frac { ( y - 9 ) ^ { 2 } } { 64 } = 1$
C) $\frac { ( x - 6 ) ^ { 2 } } { 64 } + \frac { ( y - 9 ) ^ { 2 } } { 9 } = 1$
D) $\frac { ( x - 9 ) ^ { 2 } } { 64 } + \frac { ( y - 6 ) ^ { 2 } } { 9 } = 1$
E) $\frac { ( x + 9 ) ^ { 2 } } { 64 } + \frac { ( y + 6 ) ^ { 2 } } { 9 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is at (6, 9), the midpoint of the vertices. The major axis is vertical with length 16, so $b^2 = 64$. The minor axis is horizontal with length 3, so $a^2 = 9$. The standard form is $\frac{(x-6)^2}{9} + \frac{(y-9)^2}{64} = 1$.

Find the Standard Form of the Equation of the Ellipse (x+6)29+(y+9)264=1\frac { ( x + 6 ) ^ { 2 } } { 9 } + \frac { ( y + 9 ) ^ { 2 } } { 64 } = 19(x+6)2​+64(y+9)2​=1

Find the Standard Form of the Equation of the Ellipse $\frac { ( x + 6 ) ^ { 2 } } { 9 } + \frac { ( y + 9 ) ^ { 2 } } { 64 } = 1$