Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.
Focies: (±7,0); major axis of length 16

A) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 15 } = 1$
B) $\frac { x ^ { 2 } } { 64 } - \frac { y ^ { 2 } } { 15 } = 1$
C) $- \frac { x ^ { 2 } } { 15 } + \frac { y ^ { 2 } } { 64 } = 1$
D) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 15 } = - 1$
E) $\frac { x ^ { 2 } } { 15 } + \frac { y ^ { 2 } } { 64 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. 
Focies: (±7,0); major axis of length 16

A) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 15 } = 1$
B) $\frac { x ^ { 2 } } { 64 } - \frac { y ^ { 2 } } { 15 } = 1$
C) $- \frac { x ^ { 2 } } { 15 } + \frac { y ^ { 2 } } { 64 } = 1$
D) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 15 } = - 1$
E) $\frac { x ^ { 2 } } { 15 } + \frac { y ^ { 2 } } { 64 } = 1$

Quizplus · Accepted Answer

The standard form of an ellipse with the center 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 between the center and a focus (c) is given by $c = 7$ (since the foci are at $(±7,0)$). The major axis length is 16, so $a = 8$ (half of 16). We use the relationship $c^2 = a^2 - b^2$ to find $b^2$. Plugging in the values, we get $49 = 64 - b^2$, which gives $b^2 = 15$. Therefore, the equation is $\frac{x^2}{64} + \frac{y^2}{15} = 1$.

Find the Standard Form of the Equation of the Ellipse x264+y215=1\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 15 } = 164x2​+15y2​=1

Find the Standard Form of the Equation of the Ellipse $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 15 } = 1$