Find an equation for the ellipse described.
-Center (0, 0); major axis horizontal with length 8; length of minor axis is 4
A) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$
B) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$
C) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$
D) $\frac { x ^ { 2 } } { 8 } + \frac { y ^ { 2 } } { 4 } = 1$

Question

Find an equation for the ellipse described.
-Center (0, 0); major axis horizontal with length 8; length of minor axis is 4 
A) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$
B) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$
C) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$
D) $\frac { x ^ { 2 } } { 8 } + \frac { y ^ { 2 } } { 4 } = 1$

Quizplus · Accepted Answer

The lengths of the major and minor axes are given as 8 and 4, respectively. For an ellipse centered at the origin (0,0) with a horizontal major axis, the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $2a$ is the length of the major axis and $2b$ is the length of the minor axis. Here, $2a = 8$ so $a = 4$ and $2b = 4$ so $b = 2$. Thus, $a^2 = 16$ and $b^2 = 4$, making the correct equation $\frac{x^2}{16} + \frac{y^2}{4} = 1$.

Find an Equation for the Ellipse Described x264+y216=1\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 164x2​+16y2​=1

Find an Equation for the Ellipse Described $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$