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Area of trapezium

  1. Sep 15, 2012 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    A trapezium is inscribed in the parabola y[itex]^{2}[/itex]=4x such that its diagonal pass through the point (1,0) and each has length 25/4. Find the area of trapezium.


    2. Relevant equations
    Length of chord is given by
    (4/m[itex]^{2}[/itex])([itex]\sqrt{1+m^{2}}[/itex][itex]\sqrt{a(a-mc)}[/itex]
    Here a=1 and m is an unknown quantity

    Equation of focal chord
    (t[itex]_{1}[/itex]+t[itex]_{2}[/itex])y=2(x-1)

    3. The attempt at a solution
    The length of the chord is given. Also from the equation of focal chord c=-2/t[itex]_{1}[/itex]+t[itex]_{2}[/itex] which is m. So in my first equation the only unknown quantity is m and equating it to (25/4) yields a biquadratic equation. Now finding m from here is very difficult and laborious too as the roots are not in whole numbers. Also if I find m with much difficulty(I will need a computer to get the roots lol) I have one more step to go which is to find the 4 points which I assume will take a lot of time. But nevertheless if anyhow I manage to find those 4 points I still have one more step to go before I reach the final answer and that is calculation of area which is just impossible to calculate manually. There must be some other and easier way to do this because if it would have been this difficult no one would solve it.
     
  2. jcsd
  3. Sep 15, 2012 #2

    Mentallic

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    Homework Helper

    I answered it myself and the numbers were nice. No computers needed :wink:

    Just start with equating y=m(x-1) with y2=4x then after using the quadratic formula you'll get

    [tex]x=\frac{m^2+2\pm2\sqrt{m^2+1}}{m^2}[/tex]

    then find the first (x,y) coordinates by taking the positve of the [itex]\pm[/itex] operator, then the second coordinates by taking the negative. You'll have two coordinates which represent the intersection of the diagonal through (1,0) with the parabola.

    Now since you know those, you can find the length of the chord using the distance formula.

    For that, I was able to simplify the expression down to
    [tex]\ell = \frac{4(m^2+1)}{m^2}[/tex]

    And then of course you equate this to 25/4, which will give you the values of m. Plug these back into your intercept coordinates, then you can go ahead and find the h, a and b values to plug into the trapezium formula
    [tex]A=\frac{h}{2}(a+b)[/tex]

    All good?
     
  4. Sep 16, 2012 #3

    utkarshakash

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    Hey you are great. Thank you! I got my answer within minutes. Now I feel how dumb I was as I thought this would require the use of a computer :rofl: :tongue2:
     
  5. Sep 16, 2012 #4

    Mentallic

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    If this question was given to you from class and you didn't make it up yourself, then 99% of the time you'll be able to solve it without the use of a computer. The other 1% of the time they'll tell you to solve it numerically.
    So if there was no mention of solving it numerically, you can be sure you probably just made a mistake somewhere :tongue:
     
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