Find the standard form of the equation of the ellipse centered at the origin having vertices at $( - 4,0 )$ and $( 4,0 )$ and foci at $( - 1,0 )$ and $( 1,0 )$.
A) $\frac { x ^ { 2 } } { ( \sqrt { 15 } ) ^ { 2 } } + \frac { y ^ { 2 } } { 4 ^ { 2 } } = 1$
B) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { \sqrt { 15 } } = 1$
C) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { \sqrt { 17 } } = 1$
D) $\frac { x ^ { 2 } } { \sqrt { 15 } } + \frac { y ^ { 2 } } { 4 } = 1$
E) $\frac { x ^ { 2 } } { 4 ^ { 2 } } + \frac { y ^ { 2 } } { ( \sqrt { 15 } ) ^ { 2 } } = 1$

Question

Quizplus · Accepted Answer

The standard form 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. For this ellipse, the vertices at $(-4,0)$ and $(4,0)$ indicate that the major axis length is $2a = 8$, so $a = 4$. The foci at $(-1,0)$ and $(1,0)$ indicate that the distance from the center to each focus, $c$, is 1. The relationship between $a$, $b$, and $c$ in an ellipse is $c^2 = a^2 - b^2$. Plugging in the known values gives $1^2 = 4^2 - b^2$, leading to $b^2 = 15$. Therefore, the correct equation is $\frac{x^2}{4^2} + \frac{y^2}{(\sqrt{15})^2} = 1$.

Find the Standard Form of the Equation of the Ellipse (−4,0)( - 4,0 )(−4,0)

Find the Standard Form of the Equation of the Ellipse $( - 4,0 )$