Find the center, vertices and foci of the hyperbola. $$\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 1$$
A) Center: $( 0,0 )$
Vertices: $( \pm 5,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
B) Center: $( 0,0 )$
Vertices: $\quad ( \pm 4,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
C) Center: $( 16,25 )$
Vertices: $( \pm 4,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
D) Center: $( 16,25 )$
Vertices: $( \pm 5,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
E) Center: $( 0,0 )$
Vertices: $\quad ( 0 , \pm 4 )$
Foci: $( 0 , \pm \sqrt { 41 } )$

Question

Find the center, vertices and foci of the hyperbola. $$\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 1$$ 
A) Center: $( 0,0 )$
Vertices: $( \pm 5,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
B) Center: $( 0,0 )$
Vertices: $\quad ( \pm 4,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
C) Center: $( 16,25 )$
Vertices: $( \pm 4,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
D) Center: $( 16,25 )$
Vertices: $( \pm 5,0 )$
Foci: $( \pm \sqrt { 41 } , 0 )$
E) Center: $( 0,0 )$
Vertices: $\quad ( 0 , \pm 4 )$
Foci: $( 0 , \pm \sqrt { 41 } )$

Quizplus · Accepted Answer

We can rewrite the given equation as $\dfrac{(x-0)^2}{5^2} - \dfrac{(y-0)^2}{4^2} = 1$, which is in the standard form of a hyperbola:
$\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} =1$, with center at $(h,k)$, vertices at $(h\pm a,k)$, and foci at $(h\pm c,k)$, where $c=\sqrt{a^2+b^2}$.

Comparing the given equation with the standard form, we can see that $h=0$, $k=0$, $a=5$, and $b=4$. Therefore, the center is $(0,0)$, the vertices are $(\pm 5,0)$, and $c=\sqrt{a^2+b^2}=\sqrt{5^2+4^2}=\sqrt{41}$, which gives the foci as $(\pm \sqrt{41},0)$. Hence, the correct choice is $\boxed{	extbf{(A)}}$.

Find the Center, Vertices and Foci of the Hyperbola x225−y216=1\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 125x2​−16y2​=1

Find the Center, Vertices and Foci of the Hyperbola $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 1$