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FoolForMath

Fool's problem of the day, \(1.\)Find the locus of the intersection of the perpendicular tangents drawn to the curve \(4y^3=27x^2 \). Good luck!

  • one year ago
  • one year ago

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  1. FoolForMath
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    This is somewhat difficult, took me some time to realize the solution.

    • one year ago
  2. rebeccaskell94
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    Where does one start to find the answer to this question ? you can msg me so you don't give it away if you want, I just wanna know HOW to solve it.

    • one year ago
  3. rebeccaskell94
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    Cheater. ;P

    • one year ago
  4. experimentX
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    y=3 ???

    • one year ago
  5. FoolForMath
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    No.

    • one year ago
  6. pre-algebra
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    Darn I need to read up on English definitions.ъ :(

    • one year ago
  7. ZhangYan
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    man, this is hard questions!!!

    • one year ago
  8. inkyvoyd
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    @FoolForMath , what do you mean by perp tangent/

    • one year ago
  9. inkyvoyd
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    @FoolForMath ?

    • one year ago
  10. asnaseer
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    is it x=0?

    • one year ago
  11. FoolForMath
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    Sorry, that's not the right answer.

    • one year ago
  12. ladiesman123
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    teehee :)

    • one year ago
  13. blockcolder
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    When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?

    • one year ago
  14. cinar
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    http://www.wolframalpha.com/input/?i=plot+4y^3%3D27x^2

    • one year ago
  15. FoolForMath
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    Yes, @blockcolder

    • one year ago
  16. eliassaab
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    If you try to do the easy problem for y = x^2, you find that the locus is the line y=-1/4.

    • one year ago
  17. blockcolder
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    Since the graph is symmetrical w.r.t. the y-axis, I'll consider first x>0. By implicit differentiation, we have: \[4y^3=27x^2\\ 12y^2y'=54x\\ y'=\frac{54x}{12y^2}=\frac{9x}{2y^2}\] Am I on the right track?

    • one year ago
  18. FoolForMath
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    Why do you Google Cinar? I am well aware of the answer.

    • one year ago
  19. FoolForMath
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    I am removing the spoiler.

    • one year ago
  20. FoolForMath
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    @blockcolder: looks good.

    • one year ago
  21. dumbcow
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    the set of points that represent the intercepts of the perpendicular tangents match the original function. all the intercepts lie on the same curve it was a little bit of work so i won't post my solution here...the steps were 1) find dy/dx, then the perpendicular slope is -dx/dy 2) solve for line equations for each slope 3) set them equal to find intercept points

    • one year ago
  22. asnaseer
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    ok - I misunderstood the question. I took it to mean the intersection of the lines that are perpendicular to the tangents to the given curve.

    • one year ago
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