Solve the problem.
-The cross section of a cooling tower of a nuclear power plant is in the shape of a hyperbola, and can be modeled by the equation $\frac { x ^ { 2 } } { 576 } - \frac { ( y - 83 ) ^ { 2 } } { 2,304 } = 1$ where x and y are measured in meters. The base of the tower is located at y = 0, and the top of the
Tower is 110 m above the base. Determine the diameter of the tower at the top. Round to the nearest
Meter.
A) 55 m
B) 48 m
C) 28 m
D) 96 m

Question

Quizplus · Accepted Answer

The diameter of the tower at the top is the distance between the two points where the hyperbola intersects the line y = 110. Substituting y = 110 into the equation of the hyperbola, we get:$\frac { x ^ { 2 } } { 576 } - \frac { ( 110 - 83 ) ^ { 2 } } { 2,304 } = 1$$\frac { x ^ { 2 } } { 576 } - \frac { 27 ^ { 2 } } { 2,304 } = 1$$\frac { x ^ { 2 } } { 576 } - \frac { 729 } { 2,304 } = 1$$\frac { x ^ { 2 } } { 576 } = 1 + \frac { 729 } { 2,304 } = \frac { 3,033 } { 2,304 }$$x ^ { 2 } = 576 \cdot \frac { 3,033 } { 2,304 } = 784$$x = \sqrt { 784 } \approx 28$Therefore, the diameter of the tower at the top is 28 m.

Solve the Problem x2576−(y−83)22,304=1\frac { x ^ { 2 } } { 576 } - \frac { ( y - 83 ) ^ { 2 } } { 2,304 } = 1576x2​−2,304(y−83)2​=1

Solve the Problem $\frac { x ^ { 2 } } { 576 } - \frac { ( y - 83 ) ^ { 2 } } { 2,304 } = 1$