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Arch involving CONICS hyperbola equation

  1. Apr 12, 2005 #1
    Your task is to design a curved arch similar to the a tunnel for cars. with a horizontal span of 100 m and a maximum height of 20 m.

    Using a domain of {x:-50<=x<=50} and {y:0<=y<=20} determine the following types of equations that could be used to model the curved arch.

    the equation of a hyperbola in the form [tex] \frac {(x-h)^2} {a^2} - \frac {(y-k)^2} {b^2} = -1 [/tex] where b=10 and the lower arm of the hyperbola would represent the arch.

    How do i find a and the center (h,k) please help me I am struggling on this question.
    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2
    i got h=0 and y=20 because the coordinate (0,20) is on the y-axis

    a cancels out because 0/a^2 =0 so only left with k as unknown get a quadratic function and get two values for k, which value do I take? choose a point lets say (0,50) and find a, these are the steps i have so far.

    Um for k -(20-k)^2 how do u expand this (-20+k) (-20+k) or (-20+k) (20-k)?
  4. Apr 12, 2005 #3


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    I didn't follow your problem,but i can tell u for sure that

    [tex] -(20-k)^{2}=-(20-k)(20-k)=(20-k)(k-20) [/tex]

  5. Apr 12, 2005 #4
    ok I get [tex] \frac {k^2 +40 k-400} {100} =-1 [/tex] is it possible to solve for k? once I find k i can plug it into the original equation to get a and then my standard form equation for the hyperbola representing the arch will be complete.
  6. Apr 12, 2005 #5
    Sure you can solve for k. It's a simple quadratic. Multiply both sides by 100 and then use the quadratic formula.
  7. Apr 12, 2005 #6
    hold on i changed what I did

    [tex] -\frac {(20-k)^2} {100} = -1 [/tex]

    ok I cross multiplied and got

    [tex] 400-40k+k^2=-100 [/tex]
    [tex] k^2 -40k+500=0 [/tex]

    this is my quadratic using the quadratic formula i keep getting a negative under the square root why? [tex] b^2-4(a)(c) [/tex] sqrt(-400)
    teacher said that k=10 and 30 but I dont know how.
  8. Apr 12, 2005 #7


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    U can't get a negative under the square root.U should get 400.


    P.S.It's +100 in the RHS.
  9. Apr 12, 2005 #8
    it cant be +100 on the rhs because this is the equation of a hyperbola. Using the information givin im trying to find the equation of the hyperbola that could represent a arch with a span of 100 metres and maximum height of 20 metres. I was trying to solve k so that I could sub this value into the original equation and then get a then my equation in standard form for the hyperbola will be complete.
  10. Apr 12, 2005 #9


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    [tex] -\frac{(20-k)^{2}}{100}=-1 [/tex]

    please trust me that it follows

    [tex] \frac{(20-k)^{2}}{100}=1 \Rightarrow (20-k)^{2}=100 \Rightarrow k_{1}=10,k_{2}=30 [/tex]


  11. Apr 12, 2005 #10
    the 100 is positive because you did what? :cry:

    are we cross multiplying? or multiplying both sides by -100? what are we doing?

    That equation i wrote has a negative sign too did u see that? before the brackets.
  12. Apr 12, 2005 #11


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    I simplified an equality through "-1".Or,if u prefer,i multiplied both sides through the same skinny "-1".

  13. Apr 13, 2005 #12
    Leave this for your spouse, please.
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