Explanation Definition used: Equation of a Hyperbola: Center at ( 0 , 0 ) ; Transverse axis along the x-axis: An equation of the hyperbola with the center at ( 0 , 0 ) , foci at ( − c , 0 ) and ( c , 0 ) , and vertices at ( − a , 0 ) and ( a , 0 ) is x 2 a 2 − y 2 b 2 = 1 where b 2 = c 2 − a 2 . The hyperbola has a transverse axis of length 2 a along x axis and a conjugate axis at ( 0 , ± b ) of length 2 b . “ Equation of a Hyperbola: Center at ( 0 , 0 ) ; Transverse axis along the y-axis An equation of the hyperbola with the center at ( 0 , 0 ) , foci at ( 0 , − c ) and ( 0 , c ) , and vertices at ( 0 , − a ) and ( 0 , a ) is y 2 a 2 − x 2 b 2 = 1 where b 2 = c 2 − a 2 . The hyperbola has a transverse axis of length 2 a along y axis and a conjugate axis at ( ± b , 0 ) of length 2 b . The distance between two points: The formula for distance between any two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . Calculation: Let P = ( x , y ) be a point on the ellipse and ( − 2 , 4 ) and ( − 2 , − 2 ) are the vertices of the hyperbola. Given that the conjugate axis is of length 4. That is, 2 b = 4 b = 2 The length of the transverse axis obtained by taking the difference between vertices ( − 2 , 4 ) and ( − 2 , − 2 ) . d = ( − 2 + 2 ) 2 + ( − 2 − 4 ) 2 = ( − 6 ) 2 = 6 That is, 2 a = 6 a = 3 Substitute a = 3 , b = 2 in b 2 = c 2 − a 2 and obtain the value of c , 2 2 = c 2 − 3 2 c 2 = 4 + 9 c 2 = 13 c = 13 Thus, the center of the transverse axis is, ( − 2 − 2 2 , 4 − 2 2 ) = ( − 4 2 , 2 2 ) = ( − 2 , 1 ) Therefore, the foci is ( − 2 , 1 ± c ) = ( − 2 , 1 ± 13 ) . That is, F 1 = ( − 2 , 1 + 13 ) , F 1 = ( − 2 , 1 − 13 )

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Chapter 21.6, Problem 36E

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The equation of the hyperbola with vertices (−2,4) and (−2,−2), conjugate axis with length 4 and sketch the curve.

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Chapter 21.6, Problem 35E

Chapter 21.6, Problem 37E

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Chapter 21 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

Ch. 21.1 - Find the distance between (−2, 6) and (5, −3). Ch. 21.1 - Prob. 2PE Ch. 21.1 - Prob. 1E Ch. 21.1 - Prob. 2E Ch. 21.1 - Prob. 5E Ch. 21.1 - In Exercises 5–14, find the distance between the...Ch. 21.1 - Prob. 7E Ch. 21.1 - Prob. 8E Ch. 21.1 - Prob. 9E Ch. 21.1 - Prob. 10E

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The equation of the hyperbola with vertices ( − 2 , 4 ) and ( − 2 , − 2 ) , conjugate axis with length 4 and sketch the curve.

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The equation of the hyperbola with vertices (−2,4) and (−2,−2), conjugate axis with length 4 and sketch the curve.

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Chapter 21 Solutions