Find an equation for the ellipse described.
-Center at (0, 0); focus at (-5, 0); vertex at (8, 0)
A) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 39 } = 1$
B) $\frac { x ^ { 2 } } { 39 } + \frac { y ^ { 2 } } { 64 } = 1$
C) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 64 } = 1$
D) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 39 } = 1$

Question

Find an equation for the ellipse described.
-Center at (0, 0); focus at (-5, 0); vertex at (8, 0) 
A) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 39 } = 1$
B) $\frac { x ^ { 2 } } { 39 } + \frac { y ^ { 2 } } { 64 } = 1$
C) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 64 } = 1$
D) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 39 } = 1$

Quizplus · Accepted Answer

The equation of an ellipse centered at the origin 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. The distance from the center to a focus is $c$, and $a$, $b$, and $c$ are related by $c^2 = a^2 - b^2$. Here, the vertex at (8, 0) indicates that $a = 8$, so $a^2 = 64$. The focus at (-5, 0) means $c = 5$, so $c^2 = 25$. Using $c^2 = a^2 - b^2$, we get $25 = 64 - b^2$, solving for $b^2$ gives $b^2 = 39$. Therefore, the correct equation is $\frac{x^2}{64} + \frac{y^2}{39} = 1$.

Find an Equation for the Ellipse Described x264+y239=1\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 39 } = 164x2​+39y2​=1

Find an Equation for the Ellipse Described $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 39 } = 1$