Cross-section of a Nuclear cooling tower is in the shape of a hyperbola with equation `x^2/30^2 - y^2/44^2` = 1. The tower is 150 m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower

#### Solution

Equation of hyperbola is `x^2/30^2 - y^2/44^2` = 1

Given OC = `1/2` OD and CD = 150

∴ OC = 50 m and OD = 100 m

Let the Radius of top of the tower be x_{1} and bottom of the tower be x_{2}.

∴ Points A(x_{1}, 50) and B(x_{2}, 100)

Hyperbola passes through A(x_{1}, 50)

∴ `x^2/30^2 - 50^2/44^2` = 1

`x_1^2/30^2 = 1 + 50^2/44^2`

= `1 + 2500/1936`

= `(1936 + 2500)/1936`

`x_1^2 = 30/44 xx sqrt(4436)`

⇒ `x_1 = 30/44 xx sqrt(4436)`

⇒ `x_1 = 30/44 xx (66.6)`

x_{1} = 45.41 m

∴ Radius of the top = 45.41 m.

Diameter of the top = 90.82 m

Also

The hyperbola again passes through B(x_{2}, 100)

`x^2/30^2 - 100^2/44^2` = 1

`x_1^2/30^2 = 1 + 100^2/44^2`

= `(44^2 + 100^2)/44^2`

`x_2^2 = 30^2/44^2 (11936)`

`x_2 = 30/44 xx sqrt(11936)`

= `30/44 (109.25)`

x_{2} = 74.48

∴ Radius of the base = 74.48 m.

Diameter of the base = 148.96 m

∴ Diameter of the top and base of the tower are 90.82 m and 148.96 m