Identify the vertices and the foci.
-$$( x - 6 ) ^ { 2 } + \frac { y ^ { 2 } } { 25 } = 1$$
A) vertices: $( 0,5 )$ and $( 0 , - 5 )$;
foci: $( 0,2 \sqrt { 6 } )$ and $( 0 , - 2 \sqrt { 6 } )$
B) vertices: $( 6,5 )$ and $( 6 , - 5 )$;
foci: $( 6,2 )$ and $( 6 , - 2 )$
C) vertices: $( 0,5 )$ and $( 0 , - 5 )$;
foci: $( 0,2 )$ and $( 0 , - 2 )$
D) vertices: $( 6,5 )$ and $( 6 , - 5 )$;
foci: $( 6,2 \sqrt { 6 } )$ and $( 6 , - 2 \sqrt { 6 } )$

Question

Quizplus · Accepted Answer

The given equation is in the form of an ellipse centered at $(h, k)$ with the major axis along the y-axis, given by $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$, where $a^2$ is the larger denominator. Here, $h = 6$, $k = 0$, $b^2 = 1$, and $a^2 = 25$, so $a = 5$. The vertices are at $(h, k \pm a)$, which gives $(6, \pm 5)$. The foci are at $(h, k \pm c)$, where $c = \sqrt{a^2 - b^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$, giving the foci at $(6, \pm 2\sqrt{6})$.

Identify the Vertices and the Foci (x−6)2+y225=1( x - 6 ) ^ { 2 } + \frac { y ^ { 2 } } { 25 } = 1(x−6)2+25y2​=1

Identify the Vertices and the Foci $( x - 6 ) ^ { 2 } + \frac { y ^ { 2 } } { 25 } = 1$