Find the standard form of the equation of the ellipse satisfying the given conditions.
-Foci: $( - 3,0 ) , ( 3,0 )$; vertices: $( - 4,0 ) , ( 4,0 )$
A) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 7 } = 1$
B) $\frac { x ^ { 2 } } { 7 } + \frac { y ^ { 2 } } { 16 } = 1$
C) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 7 } = 1$
D) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 16 } = 1$

Question

Quizplus · Accepted Answer

The standard form of the equation of an ellipse 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 an ellipse centered at the origin with foci along the x-axis, $a$ is half the distance between the vertices, and $c$ is half the distance between the foci. Here, $a = 4$ (since the vertices are at $(-4,0)$ and $(4,0)$), so $a^2 = 16$. The distance between the foci is $6$ (from $(-3,0)$ to $(3,0)$), so $c = 3$ and $c^2 = 9$. Using the relationship $b^2 = a^2 - c^2$, we find $b^2 = 16 - 9 = 7$. Therefore, the equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{7} = 1$.

Find the Standard Form of the Equation of the Ellipse (−3,0),(3,0)( - 3,0 ) , ( 3,0 )(−3,0),(3,0)

Find the Standard Form of the Equation of the Ellipse $( - 3,0 ) , ( 3,0 )$