Find an equation for the ellipse described.
-Focus at $( - 3,0 )$; vertices at $( \pm 5,0 )$
A) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 25 } = 1$

B) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$

C) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 25 } = 1$

D) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 16 } = 1$

Question

Find an equation for the ellipse described.
-Focus at $( - 3,0 )$; vertices at $( \pm 5,0 )$
A) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 25 } = 1$

B) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$

C) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 25 } = 1$

D) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 16 } = 1$

Quizplus · Accepted Answer

The vertices at $(\pm 5,0)$ indicate that the major axis is along the x-axis and the length of the major axis is $2a = 10$, so $a = 5$ and $a^2 = 25$. The focus at $(-3,0)$ means that the distance from the center to the focus, $c = 3$, and $c^2 = 9$. The relationship between $a$, $b$, and $c$ in an ellipse is $c^2 = a^2 - b^2$. Solving for $b^2$ gives $b^2 = a^2 - c^2 = 25 - 9 = 16$. Since the major axis is along the x-axis, $a^2 = 25$ goes under $x^2$ and $b^2 = 16$ goes under $y^2$, leading to the equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

Find an Equation for the Ellipse Described (−3,0)( - 3,0 )(−3,0)

Find an Equation for the Ellipse Described $( - 3,0 )$