Answer:
Step-by-step explanation:
Equation for a hyperboloid of one sheet, with center at the origen and axis along z-axis is:
(x/a)² + (y/b)² - (z/c)² = 1 (1)
We have to find a , b, and c
We can express equation (1)
(x/a)² + (y/b)² = (z/c)² + 1 (2)
Now if we cut the hyperboloid with planes parallel to xy plane we get for z = k ( K = 1 , 2 , 3 and so on ) circles of different radius
(x/a)² + (y/b)² = (k/c)² + 1
at z = k = 0 at the base of the hyperboloid d = 300 or r = 150 m
we have
(x/a)² + (y/b)² = 1
x² + y² = a² a² = (150)² a = b = 150
and x² + y² = (150)²
Now the other condition is at 200 m above the base d = 500 m r = 250 m minimum diameter then in equation (2) we have:
(x/a)² + (y/b)² = (z/c)² + 1
(1/a)² [ x² + y² ] = (z/c)² + 1
but x² + y² = r² and in this case r = 250 m then
(250)²/(150)² = (z/c)² + 1 ⇒ (62500/ 22500) = (200/c)² + 1
2,78 = 40000/c² + 1
2.78c² = 40000 + c²
1.78c² = 40000
c² = 40000/1.78
c² = 22471.91
c = 149,91
Then we finally have the equation:
x²/(150)² + y² /(150)² - z²/149,91 = 1
