Identify the vertices and asymptotes of the hyperbola. y 2 − x 2 = 4 y ^ { 2 } - x ^ { 2 } = 4 y 2 − x 2 = 4 A) vertices: ( 0 , − 2 ) , ( 0 , 2 ) ( 0 , - 2 ) , ( 0,2 ) ( 0 , − 2 ) , ( 0 , 2 ) asymptotes: y = − x , y = x y = - x , y = x y = − x , y = x B) vertices: ( − 4 , 0 ) , ( 4 , 0 ) ( - 4,0 ) , ( 4,0 ) ( − 4 , 0 ) , ( 4 , 0 ) asymptotes: y = − x , y = x y = - x , y = x y = − x , y = x C) vertices: ( − 2 , 0 ) , ( 2 , 0 ) ( - 2,0 ) , ( 2,0 ) ( − 2 , 0 ) , ( 2 , 0 ) asymptotes: y = − x , y = x y = - x , y = x y = − x , y = x D) vertices: ( − 2 , 2 ) , ( 2 , 2 ) ( - 2,2 ) , ( 2,2 ) ( − 2 , 2 ) , ( 2 , 2 ) asymptotes: y = − 2 x , y = x 2 y = - 2 x , y = \frac { x } { 2 } y = − 2 x , y = 2 x E) vertices: ( 0 , − 2 ) , ( 0 , 2 ) ( 0 , - 2 ) , ( 0,2 ) ( 0 , − 2 ) , ( 0 , 2 ) asymptotes: y = − 2 x , y = x 2 y = - 2 x , y = \frac { x } { 2 } y = − 2 x , y = 2 x

Question

Identify the vertices and asymptotes of the hyperbola.   y 2 − x 2 = 4 y ^ { 2 } - x ^ { 2 } = 4 y 2 − x 2 = 4 A) vertices:   ( 0 , − 2 )  , ( 0 , 2 )  ( 0 , - 2 )  , ( 0,2 )  ( 0 , − 2 )  , ( 0 , 2 )    asymptotes:   y = − x , y = x y = - x , y = x y = − x , y = x B) vertices:   ( − 4 , 0 )  , ( 4 , 0 )  ( - 4,0 )  , ( 4,0 )  ( − 4 , 0 )  , ( 4 , 0 )    asymptotes:   y = − x , y = x y = - x , y = x y = − x , y = x C) vertices:   ( − 2 , 0 )  , ( 2 , 0 )  ( - 2,0 )  , ( 2,0 )  ( − 2 , 0 )  , ( 2 , 0 )    asymptotes:   y = − x , y = x y = - x , y = x y = − x , y = x D) vertices:   ( − 2 , 2 )  , ( 2 , 2 )  ( - 2,2 )  , ( 2,2 )  ( − 2 , 2 )  , ( 2 , 2 )    asymptotes:   y = − 2 x , y = x 2 y = - 2 x , y = \frac { x } { 2 } y = − 2 x , y = 2 x ​ E) vertices:   ( 0 , − 2 )  , ( 0 , 2 )  ( 0 , - 2 )  , ( 0,2 )  ( 0 , − 2 )  , ( 0 , 2 )    asymptotes:   y = − 2 x , y = x 2 y = - 2 x , y = \frac { x } { 2 } y = − 2 x , y = 2 x ​

Marian Mante · Accepted Answer

Final Answer : A, Explanation :   The given equation   y 2 − x 2 = 4 y^2 - x^2 = 4 y 2 − x 2 = 4   can be rewritten in standard form as   y 2 2 2 − x 2 2 2 = 1 \frac{y^2}{2^2} - \frac{x^2}{2^2} = 1 2 2 y 2 ​ − 2 2 x 2 ​ = 1  , indicating a hyperbola centered at the origin (0,0) with vertices along the y-axis at   ( 0 , ± 2 ) (0, \pm2) ( 0 , ± 2 )   and asymptotes given by   y = ± x y = \pm x y = ± x  , matching choice A.

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