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anonymous

  • one year ago

A railroad tunnel is shaped like a semiellipse as shown below. The height of the tunnel at the center is 58 ft and the vertical clearance must be 29 ft at a point 21 ft from the center. Find an equation for the ellipse.

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  1. anonymous
    • one year ago
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    Diagram: http://prntscr.com/7mjbvf

  2. jim_thompson5910
    • one year ago
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    not sure, but let me think

  3. anonymous
    • one year ago
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    Anything?

  4. jim_thompson5910
    • one year ago
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    is there anything about the ellipse not mentioned? I feel like there's some missing info

  5. anonymous
    • one year ago
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    nope

  6. anonymous
    • one year ago
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    vertex is 0,58 and there are two points on the ellipse at +-21,29

  7. jim_thompson5910
    • one year ago
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    if the ellipse is taller than it is wide, then a = 58 and this pairs up with the y^2 term

  8. jim_thompson5910
    • one year ago
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    and if (21,29) is on the ellipse, then (x,y) = (21,29) --> x = 21 and y = 29

  9. jim_thompson5910
    • one year ago
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    assume the center is (h,k) = (0,0)

  10. jim_thompson5910
    • one year ago
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    \[\Large \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2} = 1\] \[\Large \frac{(21-0)^2}{b^2}+\frac{(29-0)^2}{58^2} = 1\] solve for b

  11. anonymous
    • one year ago
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    24.25ish

  12. jim_thompson5910
    • one year ago
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    actually you don't have to solve for b you can stop at b^2

  13. anonymous
    • one year ago
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    well f***

  14. jim_thompson5910
    • one year ago
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    since b^2 is in the denominator under the x^2 term

  15. anonymous
    • one year ago
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    I'm getting a "Max Iterations Error" on my calculator

  16. jim_thompson5910
    • one year ago
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    what kind of calculator do you have?

  17. anonymous
    • one year ago
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    TI-36x Pro

  18. jim_thompson5910
    • one year ago
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    I'm not familiar with that type

  19. jim_thompson5910
    • one year ago
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    but you should get b^2 = 588

  20. anonymous
    • one year ago
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    It has a number solve function where I can type in 2 sides of an equation and It'll solve for a variable in the equation. That's what I was using and it gave that error. When I solved for b^2 it works

  21. anonymous
    • one year ago
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    ok

  22. jim_thompson5910
    • one year ago
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    hmm strange

  23. anonymous
    • one year ago
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    ok so we found b^2 and b, so now what?

  24. jim_thompson5910
    • one year ago
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    replace b^2 with 588 and a^2 with whatever 58^2 is

  25. jim_thompson5910
    • one year ago
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    (h,k) = (0,0)

  26. jim_thompson5910
    • one year ago
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    x and y are left alone in the equation

  27. anonymous
    • one year ago
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    so \[\large \frac{ x^2 }{ 588 }+\frac{ y^2 }{ 3364 } =1\]

  28. anonymous
    • one year ago
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    then we need to add the values

  29. anonymous
    • one year ago
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    @jim_thompson5910

  30. jim_thompson5910
    • one year ago
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    add values?

  31. anonymous
    • one year ago
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    is that the equation or is there something left?

  32. jim_thompson5910
    • one year ago
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    the last thing you posted is the equation they want

  33. jim_thompson5910
    • one year ago
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    I guess you could solve for y to get some expression in the form y = ... that will get you the top half of the ellipse, which is the tunnel

  34. anonymous
    • one year ago
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    should I do that?

  35. jim_thompson5910
    • one year ago
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    hmm now I'm not sure if they want the full ellipse or just the upper half ellipse

  36. anonymous
    • one year ago
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    ill give both

  37. jim_thompson5910
    • one year ago
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    good idea

  38. anonymous
    • one year ago
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    So do I just solve for y=?

  39. jim_thompson5910
    • one year ago
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    yeah

  40. anonymous
    • one year ago
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    So I'm stuck at \[\frac{ y^2 }{ 3364 }=1-\frac{ x^2 }{ 588 }\]

  41. jim_thompson5910
    • one year ago
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    multiply both sides by 3364 then take the square root of both sides you focus on the positive square root because you want the upper half

  42. anonymous
    • one year ago
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    how I multiply x^2/588 by 3364 and what would that come out to

  43. jim_thompson5910
    • one year ago
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    \[\Large \frac{ y^2 }{ 3364 }=1-\frac{ x^2 }{ 588 }\] \[\Large 3364*\frac{ y^2 }{ 3364 }=3364*(1-\frac{ x^2 }{ 588 })\] \[\Large y^2=3364*(1-\frac{ x^2 }{ 588 })\] \[\Large y^2=3364*1-3364*\frac{ x^2 }{ 588 }\] \[\Large y^2=3364-\frac{3364x^2 }{ 588 }\] \[\Large y^2=3364-\frac{841x^2 }{ 147 }\] \[\Large y=???\]

  44. anonymous
    • one year ago
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    It was that last part I was confused about.\[y=58-\frac{ 29x }{ 7\sqrt{3} }\]

  45. jim_thompson5910
    • one year ago
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    you can't take the square root like that

  46. anonymous
    • one year ago
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    well crap

  47. jim_thompson5910
    • one year ago
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    \[\Large y^2=3364-\frac{841x^2 }{ 147 }\] \[\Large \sqrt{y^2}=\sqrt{3364-\frac{841x^2 }{ 147 }}\] \[\Large y=\sqrt{3364-\frac{841x^2 }{ 147 }}\]

  48. jim_thompson5910
    • one year ago
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    you apply the square root to the entire side

  49. anonymous
    • one year ago
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    then what? Is that all we can do

  50. jim_thompson5910
    • one year ago
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    yeah that's as far as you can go

  51. anonymous
    • one year ago
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    Great thanks

  52. jim_thompson5910
    • one year ago
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    \[\Large \sqrt{x + y} \ne \sqrt{x} + \sqrt{y}\]

  53. anonymous
    • one year ago
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    It's been a long day of math for me. Finally off to bed. Thanks for all the help, dude!

  54. jim_thompson5910
    • one year ago
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    no problem

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