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Find the Points on an Ellipse Furthest Away From (1,0)

  1. Apr 24, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find the points on the ellipse 4x2+y2=4 that are furthest from the point (1,0) on the ellipse.

    2. Relevant equations
    Ellipse: y=±√(4-4x2)
    Distance Formula: d=√[(x2-x1)2+(y2-y1)2]

    3. The attempt at a solution
    The distance from (1,0) for any point on the ellipse should be d=√[(x-1)2+(±√(4-4x2)2].
    That simplifies to d=√[(x-1)2+4-4x2]. The ± in front of √(4-4x2) just turns positive because the final value is squared anyways, right?

    However, when I graph the distance formula it doesn't match up with my ellipse. According to my solution manual the maximum distance occurs at x=-1/3, but my distance formula graph just keeps increasing. The manual squares d to give s=d2 and then graphs s, which does give a maximum value for s at x=-1/3, but I don't understand why they do this and how it works.

    Edit: Added negative sign to x=-1/3.
     
    Last edited: Apr 24, 2015
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  3. Apr 24, 2015 #2

    Ray Vickson

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    You should show your work, because I am in agreement with your solution manual. Presumably, you have made an error somewhere.
     
  4. Apr 24, 2015 #3

    Drakkith

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    I thought I did so. What else would you like to see?
     
  5. Apr 24, 2015 #4

    vela

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    I'm not in agreement with your solution manual. I find the maxima occur at ##x=-1/3##. If you plot the ellipse, you should be able to see x=+1/3 doesn't make sense either.
     
  6. Apr 24, 2015 #5

    Drakkith

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    Whoops, I forgot to put the negative sign in front of the 1/3. You are correct, Vela, it is -1/3.
     
  7. Apr 24, 2015 #6

    vela

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    The only thing I can think of is that you're not plotting the right function. What you wrote in the OP is fine. Whether you plot ##d## or ##d^2##, you should see a maximum at x=-1/3.
     
  8. Apr 24, 2015 #7
    Another approach that I find much simpler is to take the derivative of the function ##d^2## w.r.t ##x## and let that equal to 0 to find the maxima points.
    since the maximum power of ##x## in ##d^2## is 2, taking the derivative will produce a linear equation and you should have your desired solution.
    [EDIT:- @Drakkith you being a mentor why does your profile have the label STARTER on it???]
     
  9. Apr 24, 2015 #8

    Drakkith

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    Dang it, you are correct. I put 4-4X instead of 4-4X2 into my function. Now it has a maximum at x=-1/3.
    I swear... I checked my work about 5 times and still missed that. Thanks, Vela.

    Thanks. I'll try that. I suppose it would be a good way to double check my answer.


    Because I started this thread.
     
  10. Apr 24, 2015 #9
    Oh I see.... I always thought a starter was someone who was had just created his PF account a week earlier or something like that................
     
  11. Apr 24, 2015 #10

    Drakkith

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    Nope. It's just an easy way to see who started the thread. I'm really happy Greg added it.
     
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