Find the center, vertices and foci of the hyperbola. $5 x ^ { 2 } - 6 y ^ { 2 } = 30$
A)Center: (1, 1)Vertices: $( 0 , \pm \sqrt { 6 } )$ Foci: $( 0 , \pm \sqrt { 11 } )$
B)Center: $( 5,6 )$ Vertices: $( \pm \sqrt { 6 } , 0 )$ Foci: $( \pm \sqrt { 11 } , 0 )$
C)Center: (0, 0)Vertices: $( 0 , \pm \sqrt { 6 } )$ Foci: $( 0 , \pm \sqrt { 11 } )$
D)Center: (0, 0)Vertices: $( \pm \sqrt { 6 } , 0 )$ Foci: $( \pm \sqrt { 11 } , 0 )$
E)Center: $( 6,5 )$ Vertices: $( \pm \sqrt { 6 } , 0 )$ Foci: $( \pm \sqrt { 11 } , 0 )$

Question

Find the center, vertices and foci of the hyperbola.  $5 x ^ { 2 } - 6 y ^ { 2 } = 30$ 
A)Center: (1, 1)Vertices: $( 0 , \pm \sqrt { 6 } )$ Foci: $( 0 , \pm \sqrt { 11 } )$
B)Center: $( 5,6 )$ Vertices: $( \pm \sqrt { 6 } , 0 )$ Foci: $( \pm \sqrt { 11 } , 0 )$
C)Center: (0, 0)Vertices: $( 0 , \pm \sqrt { 6 } )$ Foci: $( 0 , \pm \sqrt { 11 } )$
D)Center: (0, 0)Vertices: $( \pm \sqrt { 6 } , 0 )$ Foci: $( \pm \sqrt { 11 } , 0 )$
E)Center: $( 6,5 )$ Vertices: $( \pm \sqrt { 6 } , 0 )$ Foci: $( \pm \sqrt { 11 } , 0 )$

Quizplus · Accepted Answer

The equation $5x^2 - 6y^2 = 30$ can be rewritten as $\frac{x^2}{6} - \frac{y^2}{5} = 1$, indicating a hyperbola centered at the origin (0, 0) with its major axis along the x-axis. The vertices are found at $\pm \sqrt{6}$ on the x-axis, and the foci are found using the formula $c^2 = a^2 + b^2$, giving $\pm \sqrt{11}$ on the x-axis.

Find the Center, Vertices and Foci of the Hyperbola 5x2−6y2=305 x ^ { 2 } - 6 y ^ { 2 } = 305x2−6y2=30

Find the Center, Vertices and Foci of the Hyperbola $5 x ^ { 2 } - 6 y ^ { 2 } = 30$