Graph the ellipse and locate the foci.
-$$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 25 } = 1$$

A) foci at $( 2 \sqrt { 14 } , 0 )$ and $( - 2 \sqrt { 14 } , 0 )$
B) foci at $( 0,2 \sqrt { 14 } )$ and $( 0 , - 2 \sqrt { 14 } )$
C) foci at $( 5 \sqrt { 3 } , 0 )$ and $( - 5 \sqrt { 3 } , 0 )$
D) foci at $( 0,5 \sqrt { 3 } )$ and $( 0 , - 5 \sqrt { 3 } )$

Question

Graph the ellipse and locate the foci.
-$$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 25 } = 1$$

A) foci at $( 2 \sqrt { 14 } , 0 )$ and $( - 2 \sqrt { 14 } , 0 )$ 
B) foci at $( 0,2 \sqrt { 14 } )$ and $( 0 , - 2 \sqrt { 14 } )$ 
C) foci at $( 5 \sqrt { 3 } , 0 )$ and $( - 5 \sqrt { 3 } , 0 )$ 
D) foci at $( 0,5 \sqrt { 3 } )$ and $( 0 , - 5 \sqrt { 3 } )$

Quizplus · Accepted Answer

The equation is in standard form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ where $(h,k)$ is the center, $a$ is the length of the semi-major axis, and $b$ is the length of the semi-minor axis. Therefore, the center is $(0,0)$, the semi-major axis is 9 and the semi-minor axis is 5. The distance from the center to each focus is $\sqrt{a^2-b^2}=\sqrt{81-25}=4\sqrt{14}$. Since the foci lie on the x-axis, the foci are at $(\pm4\sqrt{14},0)$. Simplifying this expression, we get the answer choice A.

Graph the Ellipse and Locate the Foci x281+y225=1\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 25 } = 181x2​+25y2​=1

Graph the Ellipse and Locate the Foci $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 25 } = 1$