Find an equation for the ellipse described.
-Foci at (1, 4)and (7, 4); length of major axis is 10
A) $\frac { ( x + 4 ) ^ { 2 } } { 25 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( y + 4 ) ^ { 2 } } { 25 } + \frac { ( x - 4 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x - 4 ) ^ { 2 } } { 25 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( x - 4 ) ^ { 2 } } { 25 } + \frac { ( x + 4 ) ^ { 2 } } { 16 } = 1$

Question

Find an equation for the ellipse described.
-Foci at (1, 4)and (7, 4); length of major axis is 10 
A) $\frac { ( x + 4 ) ^ { 2 } } { 25 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( y + 4 ) ^ { 2 } } { 25 } + \frac { ( x - 4 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x - 4 ) ^ { 2 } } { 25 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( x - 4 ) ^ { 2 } } { 25 } + \frac { ( x + 4 ) ^ { 2 } } { 16 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of the foci, which is $(\frac{1+7}{2}, \frac{4+4}{2}) = (4, 4)$. The major axis length is 10, so the semi-major axis length is $\frac{10}{2} = 5$, which gives $a^2 = 25$. Since the foci and the major axis are horizontal, $a^2$ goes under $(x - h)^2$, and $b^2$ will be under $(y - k)^2$. The distance between the foci is 6, so $c = 3$, and $c^2 = 9$. Using $c^2 = a^2 - b^2$, we find $b^2 = a^2 - c^2 = 25 - 9 = 16$. Therefore, the equation is $\frac{(x - 4)^2}{25} + \frac{(y - 4)^2}{16} = 1$.

Find an Equation for the Ellipse Described (x+4)225+(y+4)216=1\frac { ( x + 4 ) ^ { 2 } } { 25 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 125(x+4)2​+16(y+4)2​=1

Find an Equation for the Ellipse Described $\frac { ( x + 4 ) ^ { 2 } } { 25 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$