Find the center,vertices and foci of the hyperbola. $\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 1$
A)Center: (0,0) Vertices: $( 0 , \pm 1 )$
Foci: $( 0 , \pm \sqrt { 37 } )$
B)Center: $( 36,1 )$ Vertices: $( 0 , \pm \sqrt { 37 } )$
Foci: $( 0 , \pm 1 )$
C)Center: (0,0) Vertices: $( \pm 1,0 )$
Foci: $( \pm \sqrt { 37 } , 0 )$
D)Center: $( 36,1 )$ Vertices: $( 0 , \pm 1 )$
Foci: $( 0 , \pm \sqrt { 37 } )$
E)Center: $( 36,1 )$ Vertices: $( \pm 1,0 )$
Foci: $( \pm \sqrt { 37 } , 0 )$

Question

Find the center,vertices and foci of the hyperbola.​ $\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 1$ ​
A)Center: (0,0) Vertices: $( 0 , \pm 1 )$
Foci: $( 0 , \pm \sqrt { 37 } )$
B)Center: $( 36,1 )$ Vertices: $( 0 , \pm \sqrt { 37 } )$
Foci: $( 0 , \pm 1 )$
C)Center: (0,0) Vertices: $( \pm 1,0 )$
Foci: $( \pm \sqrt { 37 } , 0 )$
D)Center: $( 36,1 )$ Vertices: ​ $( 0 , \pm 1 )$
Foci: $( 0 , \pm \sqrt { 37 } )$
E)Center: $( 36,1 )$ Vertices: $( \pm 1,0 )$
Foci: $( \pm \sqrt { 37 } , 0 )$

Quizplus · Accepted Answer

The given equation of the hyperbola is $\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 1$. This equation is in the standard form of a hyperbola centered at the origin (0,0) with the y-axis as its transverse axis, which means the vertices are along the y-axis. The length of the transverse axis is $2a$, where $a^2 = 1$, so $a = 1$, giving the vertices as $(0, \pm 1)$. The distance from the center to the foci is $c$, where $c^2 = a^2 + b^2$, and $b^2 = 36$, so $c = \sqrt{1 + 36} = \sqrt{37}$, giving the foci as $(0, \pm \sqrt{37})$.

Find the Center,vertices and Foci of the Hyperbola y21−x236=1\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 11y2​−36x2​=1

Find the Center,vertices and Foci of the Hyperbola $\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 1$