Find the center,vertices and foci of the hyperbola. $3 x ^ { 2 } - 4 y ^ { 2 } = 12$
A)Center: (1,1) Vertices: (±4,0)
Foci: $( \pm \sqrt { 7 } , 0 )$
B)Center: $( 3,4 )$ Vertices: (±4,0)
Foci: $( \pm \sqrt { 7 } , 0 )$
C)Center: (0,0) Vertices: (0,± $2$ )
Foci: $( 0 , \pm \sqrt { 7 } )$
D)Center: (1,1) Vertices: (0,± $2$ )
Foci: $( 0 , \pm \sqrt { 7 } )$
E)Center: (0,0) Vertices: (± $2$ ,0)
Foci: $( \pm \sqrt { 7 } , 0 )$

Question

Find the center,vertices and foci of the hyperbola.​ $3 x ^ { 2 } - 4 y ^ { 2 } = 12$ ​
A)Center: (1,1) Vertices: (±4,0)
Foci: $( \pm \sqrt { 7 } , 0 )$ ​
B)Center: $( 3,4 )$ Vertices: (±4,0)
Foci: $( \pm \sqrt { 7 } , 0 )$
C)Center: (0,0) Vertices: (0,± $2$ )
Foci: $( 0 , \pm \sqrt { 7 } )$
D)Center: (1,1) Vertices: (0,± $2$ )
Foci: $( 0 , \pm \sqrt { 7 } )$
E)Center: (0,0) Vertices: (± $2$ ,0)
Foci: $( \pm \sqrt { 7 } , 0 )$

Quizplus · Accepted Answer

The equation $3x^2 - 4y^2 = 12$ can be rewritten as $\frac{x^2}{4} - \frac{y^2}{3} = 1$ by dividing both sides by 12. This is a standard form of a hyperbola centered at the origin (0,0) with vertices along the x-axis, indicating choice E is correct. The vertices are at $a = 2$, so they are (±2,0). The foci are calculated using $c^2 = a^2 + b^2$, where $a^2 = 4$ and $b^2 = 3$, so $c^2 = 7$, meaning the foci are at $(\pm \sqrt{7}, 0)$.

Find the Center,vertices and Foci of the Hyperbola 3x2−4y2=123 x ^ { 2 } - 4 y ^ { 2 } = 123x2−4y2=12

Find the Center,vertices and Foci of the Hyperbola $3 x ^ { 2 } - 4 y ^ { 2 } = 12$