Find the standard form of the equation of the ellipse with the given characteristics. foci: $( 4,3 ) , ( 4,13 )$endpoints of the major axis: $( 4,0 ) , ( 4,16 )$
A)$\frac { x + 8 ^ { 2 } } { 64 } + \frac { y + 4 ^ { 2 } } { 39 } = 1$
B) $\frac { x - 4 ^ { 2 } } { 64 } + \frac { y - 8 ^ { 2 } } { 39 } = 1$
C) $\frac { x + 4 ^ { 2 } } { 39 } + \frac { y + 8 ^ { 2 } } { 64 } = 1$
D) $\frac { x - 4 ^ { 2 } } { 39 } + \frac { y - 8 ^ { 2 } } { 64 } = 1$
E) $\frac { x - 8 ^ { 2 } } { 64 } + \frac { y - 4 ^ { 2 } } { 39 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics. foci: $( 4,3 ) , ( 4,13 )$endpoints of the major axis: $( 4,0 ) , ( 4,16 )$ 
A)$\frac { x + 8 ^ { 2 } } { 64 } + \frac { y + 4 ^ { 2 } } { 39 } = 1$
B) $\frac { x - 4 ^ { 2 } } { 64 } + \frac { y - 8 ^ { 2 } } { 39 } = 1$
C) $\frac { x + 4 ^ { 2 } } { 39 } + \frac { y + 8 ^ { 2 } } { 64 } = 1$
D) $\frac { x - 4 ^ { 2 } } { 39 } + \frac { y - 8 ^ { 2 } } { 64 } = 1$
E) $\frac { x - 8 ^ { 2 } } { 64 } + \frac { y - 4 ^ { 2 } } { 39 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of the major axis, which in this case is located at $(4, 8)$. Thus, the equation of the ellipse will have the form 
$$\frac{(x-4)^2}{a^2}+\frac{(y-8)^2}{b^2}=1$$
where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively. The distance between the foci is $2c=10$, and the length of the major axis is $2a=16$, so we have the relation $c^2=a^2-b^2$. Squaring both sides and plugging in the values we obtain $b^2=39$. Also, the distance between the center and each endpoint of the major axis is $a$, so we have $a=8$. Substituting these values into the equation above gives 
$$\frac{(x-4)^2}{64}+\frac{(y-8)^2}{39}=1$$

Find the Standard Form of the Equation of the Ellipse (4,3),(4,13)( 4,3 ) , ( 4,13 )(4,3),(4,13)

Find the Standard Form of the Equation of the Ellipse $( 4,3 ) , ( 4,13 )$