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

Question

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

Quizplus · Accepted Answer

The center of the hyperbola is at the origin (0, 0) because there are no shifts in the x or y directions in the equation. The vertices are found by taking the square root of the denominator under the x^2 term, which gives $\pm 4$, so the vertices are at $(\pm 4, 0)$. The foci are found by calculating $c = \sqrt{a^2 + b^2}$, where $a^2 = 16$ and $b^2 = 25$, giving $c = \sqrt{41}$, so the foci are at $(\pm \sqrt{41}, 0)$.

Find the Center, Vertices and Foci of the Hyperbola x216−y225=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 25 } = 116x2​−25y2​=1

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