Mastering Hyperbola Problems: Asymptotes and Applications

A satellite dish is shaped like a hyperbola. If the distance between the foci of the dish is 10 meters and the distance from the center to a vertex is 4 meters, what are the equations of the asymptotes?

• y = (5/4)x, y = -(5/4)x

• y = (1/2)x, y = -(1/2)x

• y = (4/3)x, y = -(4/3)x

• y = (3/4)x, y = -(3/4)x

A hyperbola is used in a cooling tower design. If the transverse axis is 12 meters and the conjugate axis is 16 meters, find the equations of the asymptotes.

• y = 4x, y = -4x

• y = (3/4)x, y = -(3/4)x

• y = (4/3)x, y = -(4/3)x

• y = (2/3)x, y = -(2/3)x

A car is traveling along a road that curves in the shape of a hyperbola. If the distance from the center to a vertex is 5 miles and the distance between the foci is 8 miles, what is the equation of the hyperbola?

• (y^2/20) - (x^2/16) = 1

• (y^2/30) - (x^2/4) = 1

• (y^2/25) - (x^2/9) = 1

• (y^2/10) - (x^2/25) = 1

An engineer is designing a bridge that resembles a hyperbola. If the distance between the foci is 14 meters and the distance from the center to a vertex is 7 meters, calculate the asymptotes of the hyperbola.

• y = 0

• y = 7

• y = -7

• y = 14

A hyperbola is formed by the path of a satellite. If the distance from the center to a vertex is 3 km and the distance between the foci is 10 km, what are the equations of the asymptotes?

• y = 2x + 1 and y = -2x + 1

• y = (4/3)x and y = -(4/3)x

• y = (3/4)x and y = -(3/4)x

• y = (5/3)x and y = -(5/3)x

A hyperbola is used to model the trajectory of a water fountain. If the transverse axis is 20 feet and the conjugate axis is 30 feet, find the equations of the asymptotes.

• y = (1/2)x and y = -(1/2)x

• y = (3/2)x and y = -(3/2)x

• y = (2/3)x and y = -(2/3)x

• y = 3x and y = -3x

A hyperbola is used in a physics experiment to study sound waves. If the distance between the foci is 12 meters and the distance from the center to a vertex is 6 meters, what is the equation of the hyperbola?

• y^2/36 - x^2/0 = 1

• y^2/36 - x^2/12 = 1

• y^2/36 + x^2/0 = 1

• y^2/24 - x^2/0 = 1

A hyperbola is used in a design for a new park. If the transverse axis is 18 meters and the conjugate axis is 24 meters, calculate the asymptotes of the hyperbola.

• y = ±(2/3)x

• y = ±(4/3)x

• y = ±(5/4)x

• y = ±(3/4)x

A hyperbola is used in a telescope design. If the distance from the center to a vertex is 4 meters and the distance between the foci is 8 meters, what are the equations of the asymptotes?

• y = -4

• y = 4

• y = 2x

• y = 0

A hyperbola is used in a new architectural design. If the transverse axis is 15 meters and the conjugate axis is 20 meters, find the equations of the asymptotes.

• y = (2/3)x and y = -(2/3)x

• y = (5/4)x and y = -(5/4)x

• y = (4/3)x and y = -(4/3)x

• y = (3/4)x and y = -(3/4)x