Find the standard form of the equation of the ellipse satisfying the given conditions.
-Endpoints of major axis: (2, -8)and (2, 0); endpoints of minor axis: (0, -4)and (4, -4);
A) $\frac { ( x - 2 ) ^ { 2 } } { 4 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( x - 2 ) ^ { 2 } } { 4 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x + 2 ) ^ { 2 } } { 4 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( x + 4 ) ^ { 2 } } { 4 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$

Question

Find the standard form of the equation of the ellipse satisfying the given conditions.
-Endpoints of major axis: (2, -8)and (2, 0); endpoints of minor axis: (0, -4)and (4, -4); 
A) $\frac { ( x - 2 ) ^ { 2 } } { 4 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( x - 2 ) ^ { 2 } } { 4 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( x + 2 ) ^ { 2 } } { 4 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( x + 4 ) ^ { 2 } } { 4 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of both the major and minor axes, which is (2, -4). The length of the major axis is 8 (from -8 to 0), so its semi-major axis is 4, and the length of the minor axis is 4 (from 0 to 4), so its semi-minor axis is 2. The standard form of the ellipse equation is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where (h, k) is the center, $a^2$ is the square of the semi-major axis, and $b^2$ is the square of the semi-minor axis. Thus, the correct equation is $\frac{(x-2)^2}{4} + \frac{(y+4)^2}{16} = 1$.

Find the Standard Form of the Equation of the Ellipse (x−2)24+(y+4)216=1\frac { ( x - 2 ) ^ { 2 } } { 4 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 14(x−2)2​+16(y+4)2​=1

Find the Standard Form of the Equation of the Ellipse $\frac { ( x - 2 ) ^ { 2 } } { 4 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$