The sides of a nuclear power plant cooling tower form a hyperbola. The diameter of the bottom of the toweis 286 feet. The smallest diameter of the tower is 168 which is 435 feet above the ground. The tower is 590feet tall.168 ft590 ft435 ft286 ftPhoto remixed from work by Jiří Sedláček r?!, cC-BY-SAFind the width of the tower at a height of 128 feet, to 1 decimal place.feet.

Given:The sides of the nuclear power plant form a hyperbola.Height of the tower = 590 ftDiameter of the bottom of the tower = 286 ft.Smallest diameter of the tower = 168 ft, which is at a height of 435 ft from the ground. We have to find the diameter of the tower at the height of 128 ft.The equation of the hyperbola is given byx2a2-y2b2=1Consider that the origin of the hyperbola is the mid-point of the diameter of hyperbola, where the diameter is smallest.The smallest diameter is 168 ft, which is equal to 2a.As, 2a = 168 ⇒ a = 84 ft, substituting this in the equation of hyperbola.x284-y2b2=1 ......(1)Using this, we can determine the value of b. Now, another co-ordinates can be understood as:The diameter of the base of the tower is 286 ft ⇒a = 143 ft.The origin (the smallest diameter) is at a height of 435 ft from the ground. So, we get, b = -435 ft.The co-ordinates of the point are (143, -435).Substituting these co-ordinates in (1), we get, (143)2(84)2-(-435)2b2=1⇒(143)2(84)2-1 =(-435)2b2⇒b2 =(-435)2(143)2(84)2-1=99691.74942⇒b= 315.74 ≈316 Thus, the equation of the hyperbola becomesx2(84)2-y2(316)2=1The distance of 128 ft from the ground is 435-128 = 307 ft away from the origin in the negative direction.Put y = -307 ft in the equation,x2(84)2-(307)2(316)2=1⇒x2(84)2=1+(307)2(316)2⇒x2 = (84)21+(307)2(316)2⇒x2=13715.79955⇒x= 117.115The value x = 117.1 ft is the radius of the tower at the height of 128 ft from the ground.The width of the tower = 2*117.1 = 234.2 ft.