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Hyperbola equations

  • Thread starter yourmom98
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  • #1
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Using a domain of {x | -50 < x < 50} and a range of {y | 0 < y < 20}, determine the following types of equations that you could use to model the curved arch.

The equation of a hyperbola in the form , where b = 10. The lower arm of the hyperbola would represent the arch.
((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=-1

i have 2 questions about this number one shouldn't the equations be

((x-h)^2)/(b^2)-((y-k)^2)/(a^2)=-1 because it should be a hyberbola that opens up and down?
and second how do i solve this?
 

Answers and Replies

  • #2
Andrew Mason
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yourmom98 said:
Using a domain of {x | -50 < x < 50} and a range of {y | 0 < y < 20}, determine the following types of equations that you could use to model the curved arch.

The equation of a hyperbola in the form , where b = 10. The lower arm of the hyperbola would represent the arch.
((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=-1

i have 2 questions about this number one shouldn't the equations be

((x-h)^2)/(b^2)-((y-k)^2)/(a^2)=-1 because it should be a hyberbola that opens up and down?
and second how do i solve this?
I think you are missing part of the problem. Does the question give you a value for C, the distance from the origin to the foci?

AM
 
  • #3
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nope it does say that the graph is supposed to be a curved arch that will have horizontal span of 100m and a maximum height of 20m.
 
  • #4
Andrew Mason
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yourmom98 said:
nope it does say that the graph is supposed to be a curved arch that will have horizontal span of 100m and a maximum height of 20m.
Ok. That is useful. Assume the curve touches the x axis at x=-50 and +50 and reaches maximum y of 20 at x=0. You just have to determine the values for a and k (h=0 if it is to be centred around x=0).

Let x=0 and y = 20 to work out value for k: [itex](y-k)^2 = b^2[/itex]
Then let x = 50 and y=0 to get the value for a: [itex]b^2x^2 - a^2(y-k)^2 = -a^2b^2[/itex]

AM
 

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