Find the vertices and foci of the ellipse.
-$$\frac { x ^ { 2 } } { 400 } + \frac { y ^ { 2 } } { 144 } = 1$$
A) Vertices: $( \pm 20,0 )$; Foci: $( \pm 12,0 )$
B) Vertices: $( \pm 20,0 )$; Foci: $( \pm 16,0 )$
C) Vertices: $( 0 , \pm 20 )$; Foci: $( 0 , \pm 16 )$
D) Vertices: $( 0 , \pm 20 )$; Foci: $( 0 , \pm 12 )$

Question

Quizplus · Accepted Answer

The ellipse is horizontal with $a^2 = 400$ and $b^2 = 144$. The vertices are $(\pm a, 0) = (\pm 20, 0)$. The foci are $(\pm c, 0)$ where $c = \sqrt{a^2 - b^2} = \sqrt{400 - 144} = \sqrt{256} = 16$.

Find the Vertices and Foci of the Ellipse x2400+y2144=1\frac { x ^ { 2 } } { 400 } + \frac { y ^ { 2 } } { 144 } = 1400x2​+144y2​=1

Find the Vertices and Foci of the Ellipse $\frac { x ^ { 2 } } { 400 } + \frac { y ^ { 2 } } { 144 } = 1$