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

Question

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

Quizplus · Accepted Answer

The equation of the hyperbola is given in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a^2 = 25$ and $b^2 = 16$. This indicates that the center of the hyperbola is at the origin (0, 0), and since $x^2$ is positive, the hyperbola opens horizontally. The vertices are located at $(\pm a, 0)$, which translates to $(\pm 5, 0)$. The foci are found using the formula $c^2 = a^2 + b^2$, which gives $c^2 = 25 + 16 = 41$, so the foci are at $(\pm \sqrt{41}, 0)$.

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 $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 1$