Write the standard form of the equation of the hyperbola centered at the origin. Vertices: ( − 2 , 0 ) , ( 2 , 0 ) ( - 2,0 ) , ( 2,0 ) ( − 2 , 0 ) , ( 2 , 0 ) Asymptotes: y = 5 x , y = − 5 x y = 5 x , y = - 5 x y = 5 x , y = − 5 x A) x 2 4 − y 2 4 / 25 = 1 \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 4 / 25 } = 1 4 x 2 − 4/25 y 2 = 1 B) x 2 4 − y 2 100 = 1 \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 100 } = 1 4 x 2 − 100 y 2 = 1 C) x 2 4 / 25 − y 2 4 = 1 \frac { x ^ { 2 } } { 4 / 25 } - \frac { y ^ { 2 } } { 4 } = 1 4/25 x 2 − 4 y 2 = 1 D) y 2 100 − x 2 4 = 1 \frac { y ^ { 2 } } { 100 } - \frac { x ^ { 2 } } { 4 } = 1 100 y 2 − 4 x 2 = 1 E) x 2 25 − y 2 4 = 1 \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1 25 x 2 − 4 y 2 = 1

Question

Write the standard form of the equation of the hyperbola centered at the origin. Vertices:   ( − 2 , 0 )  , ( 2 , 0 )  ( - 2,0 )  , ( 2,0 )  ( − 2 , 0 )  , ( 2 , 0 )    Asymptotes:   y = 5 x , y = − 5 x y = 5 x , y = - 5 x y = 5 x , y = − 5 x A)    x 2 4 − y 2 4 / 25 = 1 \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 4 / 25 } = 1 4 x 2 ​ − 4/25 y 2 ​ = 1 B)    x 2 4 − y 2 100 = 1 \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 100 } = 1 4 x 2 ​ − 100 y 2 ​ = 1 C)    x 2 4 / 25 − y 2 4 = 1 \frac { x ^ { 2 } } { 4 / 25 } - \frac { y ^ { 2 } } { 4 } = 1 4/25 x 2 ​ − 4 y 2 ​ = 1 D)    y 2 100 − x 2 4 = 1 \frac { y ^ { 2 } } { 100 } - \frac { x ^ { 2 } } { 4 } = 1 100 y 2 ​ − 4 x 2 ​ = 1 E)    x 2 25 − y 2 4 = 1 \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1 25 x 2 ​ − 4 y 2 ​ = 1

Sofia Scrivens · Accepted Answer

Final Answer : B, Explanation :   The standard form of a hyperbola centered at the origin with a horizontal transverse axis is   x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 a 2 x 2 ​ − b 2 y 2 ​ = 1  , where   2 a 2a 2 a   is the distance between the vertices, and   b 2 b^2 b 2   can be found from the equation of the asymptotes   y = ± b a x y = \pm \frac{b}{a}x y = ± a b ​ x  . Given vertices at   ( − 2 , 0 ) (-2,0) ( − 2 , 0 )   and   ( 2 , 0 ) (2,0) ( 2 , 0 )  ,   a = 2 a = 2 a = 2  , so   a 2 = 4 a^2 = 4 a 2 = 4  . The asymptotes are given as   y = ± 5 x y = \pm 5x y = ± 5 x  , implying   b a = 5 \frac{b}{a} = 5 a b ​ = 5  , so   b = 5 a = 10 b = 5a = 10 b = 5 a = 10  , and   b 2 = 100 b^2 = 100 b 2 = 100  . Thus, the correct equation is   x 2 4 − y 2 100 = 1 \frac{x^2}{4} - \frac{y^2}{100} = 1 4 x 2 ​ − 100 y 2 ​ = 1  .

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