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Math Help - Extremely Hard Max/Min Problem

  1. #1
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    Extremely Hard Max/Min Problem

    5. A chord joins any two points A and B on the parabola whose equation is y^2 = 4x. If C is the midpoint of AB, and CD is drawn parallel to the x-axis to meet the parabola at D, prove that the tangent at D is parallel to chord AB.

    I've attempted to graph the situation (attempted graph is attached), however I have no idea how to represent it with equations. What are my constraints? What maximum/minimum am I trying to solve for?

    Extremely Hard Max/Min Problem-mathquestion.jpg

    I've understood this unit up to now, though I simply can't find an answer to this question. Help would be strongly appreciated.
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  2. #2
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    Quote Originally Posted by RogueDemon View Post
    5. A chord joins any two points A and B on the parabola whose equation is y^2 = 4x. If C is the midpoint of AB, and CD is drawn parallel to the x-axis to meet the parabola at D, prove that the tangent at D is parallel to chord AB.

    I've attempted to graph the situation (attempted graph is attached), however I have no idea how to represent it with equations. What are my constraints? What maximum/minimum am I trying to solve for?

    Click image for larger version. 

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    I've understood this unit up to now, though I simply can't find an answer to this question. Help would be strongly appreciated.
    Your really don't need to maximize or minimize anything!

    First let y_1 be the y coordinate of the first point then this gives the ordered pair

    \displaystyle \left( \frac{y^2_1}{4},y_1\right)

    y_2 be the y coordinate of the and point then this gives the ordered pair

    \displaystyle \left( \frac{y^2_2}{4},y_2\right)

    So the mid point is

    \displaystyle \left( \frac{y_1^2+y^2_2}{8},\frac{y_1+y_2}{2}\right)

    and the slope is

    \displaystyle \left( \frac{y_2-y_1}{\frac{y^2_2-y_1^2}{4}}\right)=\frac{4}{y_1+y_2}

    Now the implicit derivative of the parabola is

    \displaystyle 2yy'=4 \iff y'=\frac{2}{y}

    but y at the midpoint is \displaystyle \frac{y_1+y_2}{2}
    Last edited by TheEmptySet; April 7th 2011 at 01:41 PM. Reason: typo
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  3. #3
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    Thank you very much!

    Your solution makes sense, though are you sure that max/mins aren't involved in this? I need to be sure so that I show all necessary steps on the upcoming test.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    Your really don't need to maximize or minimize anything!

    First let y_1 be the y coordinate of the first point then this gives the ordered pair

    \displaystyle \left( \frac{y^2_1}{4},y_1\right)

    Typo.
    Slope is 4/y1+y2, not 2/"".

    y_2 be the y coordinate of the and point then this gives the ordered pair

    \displaystyle \left( \frac{y^2_2}{4},y_2\right)

    So the mid point is

    \displaystyle \left( \frac{y_1^2+y^2_2}{8},\frac{y_1+y_2}{2}\right)

    and the slope is

    \displaystyle \left( \frac{y_2-y_1}{\frac{y^2_2-y_1^2}{4}}\right)=\frac{2}{y_1+y_2}

    Now the implicit derivative of the parabola is

    \displaystyle 2yy'=4 \iff y'=\frac{2}{y}

    but y at the midpoint is \displaystyle \frac{y_1+y_2}{2}
    There's a typo there i believe.
    Slope is 4/(y1+y2).
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  5. #5
    Behold, the power of SARDINES!
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    yes you are correct. I will edit above
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by RogueDemon View Post
    Thank you very much!

    Your solution makes sense, though are you sure that max/mins aren't involved in this? I need to be sure so that I show all necessary steps on the upcoming test.
    Yes I am sure. You were not asked to maximize or minimize anything. Just prove that the tangent line is parallel to a chord. I did use implicit differentiation but that was the only calculus in the whole problem
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