HELPPPPP!!! Hyperpola and Ellipse word problem

Dec 2008
11
0
(1) A hyperbolic reflector for a radiotelescope has the equation :
y^2/900-x^2/225=1.
If the reflector has a diameter of 20 feet, how deep is it (in feet)?

(2)A nuclear cooling tower is in the shape of a hyperboloid(that is, a hyperbola rotated around its conjugate axis). The equation of the hyperbola is :
x^2/5625-y^2/10000=1
If the center of the tower (the narrowest part) is 100 feet above the ground, what is the diameter (in feet) of the circular base?

(3) A semi-elliptical arch in a bridge over a river has a span of 40 feet and a clearance of 12 feet above the water in the middle of the bridge. How much distance (in feet) above the water is there 5 feet from the banks?

PLLLLEEEASSSE help set these up for me.
 

earboth

MHF Hall of Honor
Jan 2006
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(1) A hyperbolic reflector for a radiotelescope has the equation :
y^2/900-x^2/225=1.
If the reflector has a diameter of 20 feet, how deep is it (in feet)?

...
The hyperbola is opening up and down. I take the upper part of the hyperbola.

A radius of 20' correspond to a x-value of 10.

Plug in this value into the equation and solve for y. The depth is calculated by: \(\displaystyle d = y - 30 \approx 6.0555\ '\)
 

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earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
...

(2)A nuclear cooling tower is in the shape of a hyperboloid(that is, a hyperbola rotated around its conjugate axis). The equation of the hyperbola is :
x^2/5625-y^2/10000=1
If the center of the tower (the narrowest part) is 100 feet above the ground, what is the diameter (in feet) of the circular base?

...
Place the origin in the center of the tower. Then the base circle is at y = -100'. Plug in this value into the given equation and solve for x.

The diameter in question is d = 2x. I've got \(\displaystyle r \approx 212.132'\)
 

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earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
(1) ...
(3) A semi-elliptical arch in a bridge over a river has a span of 40 feet and a clearance of 12 feet above the water in the middle of the bridge. How much distance (in feet) above the water is there 5 feet from the banks?

...
Place the origin on the surface of the water and in the middle of the bridge. The semi-axes are a = 20' and b = 12. Therefore the equation of the ellipse is:

\(\displaystyle \dfrac{x^2}{20^2} + \dfrac{y^2}{12^2}=1\)

A point 5' from the bank is 15' away from the origin. Plug in x = 15 into the equation above and solve for y (The negative value is for divers only)

I've got \(\displaystyle d \approx 7.937'\)
 

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