Find the standard form of the equation of the ellipse satisfying the given conditions.
-Endpoints of major axis: (-10, -4)and (6, -4); endpoints of minor axis: (-2, -8)and (-2, 0)
A) $\frac { ( x + 2 ) ^ { 2 } } { 64 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( x + 4 ) ^ { 2 } } { 16 } + \frac { ( y + 2 ) ^ { 2 } } { 64 } = 1$
C) $\frac { ( x - 2 ) ^ { 2 } } { 64 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 0$
D) $\frac { ( x - 2 ) ^ { 2 } } { 64 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$

Question

Find the standard form of the equation of the ellipse satisfying the given conditions.
-Endpoints of major axis: (-10, -4)and (6, -4); endpoints of minor axis: (-2, -8)and (-2, 0) 
A) $\frac { ( x + 2 ) ^ { 2 } } { 64 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( x + 4 ) ^ { 2 } } { 16 } + \frac { ( y + 2 ) ^ { 2 } } { 64 } = 1$
C) $\frac { ( x - 2 ) ^ { 2 } } { 64 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 0$
D) $\frac { ( x - 2 ) ^ { 2 } } { 64 } + \frac { ( y - 4 ) ^ { 2 } } { 16 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of both the major and minor axes. For the major axis with endpoints (-10, -4) and (6, -4), the midpoint (center of the ellipse) is $\frac{-10 + 6}{2}, \frac{-4 + (-4)}{2} = (-2, -4)$. For the minor axis with endpoints (-2, -8) and (-2, 0), the midpoint is also (-2, -4), confirming the center. The length of the major axis is the distance between (-10, -4) and (6, -4), which is 16 units, so the semi-major axis length $a = \frac{16}{2} = 8$, and $a^2 = 64$. The length of the minor axis is the distance between (-2, -8) and (-2, 0), which is 8 units, so the semi-minor axis length $b = \frac{8}{2} = 4$, and $b^2 = 16$. Therefore, the standard form of the ellipse equation is $\frac{(x + 2)^2}{64} + \frac{(y + 4)^2}{16} = 1$.

Find the Standard Form of the Equation of the Ellipse (x+2)264+(y+4)216=1\frac { ( x + 2 ) ^ { 2 } } { 64 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 164(x+2)2​+16(y+4)2​=1

Find the Standard Form of the Equation of the Ellipse $\frac { ( x + 2 ) ^ { 2 } } { 64 } + \frac { ( y + 4 ) ^ { 2 } } { 16 } = 1$