Write the equation of the ellipse with foci at (2, 2), (6, 2), and b = 4, where b - semiminor axis.
A) $\frac { ( x - 4 ) ^ { 2 } } { 20 } - \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( x - 4 ) ^ { 2 } } { 20 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x - 4 ) ^ { 2 } } { 20 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = - 1$
D) $\frac { ( x + 4 ) ^ { 2 } } { 20 } + \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$
E) $\frac { ( x - 4 ) ^ { 2 } } { 16 } + \frac { ( y - 2 ) ^ { 2 } } { 20 } = 1$

Question

Write the equation of the ellipse with foci at (2, 2), (6, 2), and b = 4, where b - semiminor axis. 
A) $\frac { ( x - 4 ) ^ { 2 } } { 20 } - \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( x - 4 ) ^ { 2 } } { 20 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x - 4 ) ^ { 2 } } { 20 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = - 1$
D) $\frac { ( x + 4 ) ^ { 2 } } { 20 } + \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$
E) $\frac { ( x - 4 ) ^ { 2 } } { 16 } + \frac { ( y - 2 ) ^ { 2 } } { 20 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of the foci, which is $(\frac{2+6}{2}, \frac{2+2}{2}) = (4, 2)$. The distance between the foci is 4 units, so $2c = 4$, meaning $c = 2$. Since $b = 4$, we use the relationship $a^2 = b^2 + c^2$ to find $a^2$. Here, $a^2 = 4^2 + 2^2 = 20$. The major axis is horizontal because the foci have the same y-coordinate. Thus, the correct equation is $\frac{(x - 4)^2}{20} + \frac{(y - 2)^2}{16} = 1$.

Write the Equation of the Ellipse with Foci at (2 (x−4)220−(y−2)216=1\frac { ( x - 4 ) ^ { 2 } } { 20 } - \frac { ( y - 2 ) ^ { 2 } } { 16 } = 120(x−4)2​−16(y−2)2​=1

Write the Equation of the Ellipse with Foci at (2 $\frac { ( x - 4 ) ^ { 2 } } { 20 } - \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$