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Find perpendicular line to hyperbola without calculus

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the equation of the line perpendicular to ## x^2-y^2=1 ## at the point ## (2,\sqrt{3}) ##?

    2. Relevant equations


    3. The attempt at a solution
    I would automatically know what to do if I could use calculus, but apparently you can solve this without it. I would also know what to do if the conic was a circle (i.e. use the difference between the point given and the center of the circle). But how can I solve it in the case of a hyperbola?
     
  2. jcsd
  3. Jul 25, 2013 #2
    Well, you COULD do ##x^2-y^2=1\rightarrow\text{*COUGHCOUGHCOUGH*}\rightarrow y=-\frac{\sqrt{3}}{2}x+\frac{3\sqrt{3}}{2}##. Or, you could think about the relation between ellipses and hyperbolas. :wink:
     
  4. Jul 26, 2013 #3

    ehild

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    Perpendicular to a curve means perpendicular to its tangent at the given point. How many common point has got a curve with its tangent line?

    ehild
     
  5. Jul 26, 2013 #4

    LCKurtz

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    Maybe lots? ##y = \sin x,\, y_T = 1##.
     
  6. Jul 26, 2013 #5

    ehild

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    Maybe, with a periodic function. How can I ask the question in the correct way?

    ehild
     
  7. Jul 26, 2013 #6

    LCKurtz

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    I don't know. I'm not sure what you are trying to point out.
     
  8. Jul 26, 2013 #7
    "What is the minimum number of elements in the intersection of the set of points satisfied by a function's relation and the set of points on its tangent line in any given neighborhood to the point of tangency?"
     
  9. Jul 26, 2013 #8

    LCKurtz

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    Be thinking about this function at ##x=0## as you try to rephrase your question, whatever it is:$$
    f(x) = x^2\sin(\frac 1 x) \text{ for }x\ne 0,\, f(0) = 0$$
     
  10. Jul 26, 2013 #9

    ehild

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    Uhh.
    I did not want to give too much hint, but I see now that even LCKurtz did not understand me. :cry:
    Without calculus, one obtains the tangent line as a secant line which has only one common point with the hyperbola in the first quadrant: the points of intersections coincide.
    The equation of the straight line is y=ax+b, with √3=2a+b. Substitute for y in the equation x2-y2=1 For a single root, the discriminant is zero. The slope of the tangent line is obtained from that condition.

    ehild
     
    Last edited: Jul 26, 2013
  11. Jul 26, 2013 #10

    ehild

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    ????? Sorry, I do not understand you with my poor English. Has this thing related anyway to the OP?


    ehild
     
  12. Jul 26, 2013 #11

    LCKurtz

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    Never mind. When you asked originally "How many common point has got a curve with its tangent line?" I thought you were asking about a general property of tangent lines that could be applied specifically to this problem. And I couldn't figure out what that general property might be.

    So, no, my last example has nothing to do with the OP's question, and you can ignore it.

    And there's nothing wrong with your English. :smile:
     
  13. Jul 29, 2013 #12
    Given hyperbola [itex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 [/itex], and tangent equation [itex]y=mx+c[/itex] the relationship [itex]c^2+b^2=a^2m^2[/itex] holds. The derivation of this is explained here. This can be combined with what we know about the relationship between [itex]m[/itex] and [itex]c[/itex] by setting
    [itex]x=2[/itex] and [itex]y=\sqrt{3}[/itex] to give the required equation.
     
  14. Aug 4, 2013 #13

    NascentOxygen

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  15. Aug 4, 2013 #14

    lurflurf

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  16. Aug 4, 2013 #15

    NascentOxygen

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    The correction is much appreciated, also the erudite flair with wolframalpha. :smile:
     
  17. Aug 4, 2013 #16

    HallsofIvy

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    I believe that what is being hinted at is method of finding a tangent line to a curve that Fermat (literally preCalculus) called "ad-equation". Any (non-vertical) line through the point [itex](2, \sqrt{3})[/itex] is of the form [itex]y= m(x- 2)+ \sqrt{3}[/itex] for some m. Certainly, in order that this line be tangent to [itex]x^2- y^2= 1[/itex] at [itex](2, \sqrt{3})[/itex], they must have that point in common. Further, in order that they be tangent, x= 2 must be a double root of [itex]y= \sqrt{x^2- 1}= (m(x- 2)+ \sqrt{3})[/itex]. Squaring both sides gives the polynomial equation [itex]x^2- 1= m^2(x- 2)^2+ 2m\sqrt{3}(x- 2)+ 3[/itex]. Obviously, x= 2 satisfies that equation. What must m be so that x= 2 is a double root?
     
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