Write the standard form of an equation of the ellipse subject to the given conditions.
-Vertices: $( 6,1 )$ and $( - 12,1 )$
Foci: $( - 3 - \sqrt { 65 } , 1 )$ and $( - 3 + \sqrt { 65 } , 1 )$
A) $\frac { ( x - 3 ) ^ { 2 } } { 9 } + \frac { ( y + 1 ) ^ { 2 } } { 4 } = 1$
B) $\frac { ( x + 3 ) ^ { 2 } } { 81 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x + 3 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 4 } = 1$
D) $\frac { ( x - 3 ) ^ { 2 } } { 81 } + \frac { ( y + 1 ) ^ { 2 } } { 16 } = 1$

Question

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of the vertices, which is $(-3, 1)$. The distance between the vertices is $6 + 12 = 18$, so the major axis length is $18$, and $a = \frac{18}{2} = 9$, giving $a^2 = 81$. The distance from the center to either focus is $\sqrt{65}$, so $c = \sqrt{65}$, and $c^2 = 65$. Using $c^2 = a^2 - b^2$, we find $b^2 = 81 - 65 = 16$. The major axis is horizontal, so $a^2$ is under $x$ and $b^2$ is under $y$, leading to the equation $\frac{(x + 3)^2}{81} + \frac{(y - 1)^2}{16} = 1$.

Write the Standard Form of an Equation of the Ellipse (6,1)( 6,1 )(6,1)

Write the Standard Form of an Equation of the Ellipse $( 6,1 )$