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    could you point me in the right direction?
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    29 views and nobody helps .... 3 hours wait
    appalling service ...
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    I am clueless on this topic. Sorry I'm a recruit
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    (Original post by TeeEm)
    29 views and nobody helps .... 3 hours wait
    appalling service ...
    I am utterly disgraceful at maths but this comment made me laugh a lot. Would have been 10/10 with the appropriate stock image from your bountiful collection :lol:
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    (Original post by iEthan)
    I am utterly disgraceful at maths but this comment made me laugh a lot. Would have been 10/10 with the appropriate stock image from your bountiful collection :lol:
    do not want to over do it ....
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    (Original post by Student403)
    I am clueless on this topic. Sorry I'm a recruit
    ONLY strong FP2/FP3 ....
    you know that I have to sign your reference if you go to the States ?
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    (Original post by TeeEm)
    do not want to over do it ....
    After your results at the TSR awards tonight, I think you could afford to let your hair down (congratulations by the way, voted for you :yep:)
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    (Original post by TeeEm)
    ONLY strong FP2/FP3 ....
    you know that I have to sign your reference if you go to the States ?
    :emo: I will crack down on the two once I have finished them by Easter
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    (Original post by iEthan)
    After your results at the TSR awards tonight, I think you could afford to let your hair down (congratulations by the way, voted for you :yep:)
    I have let my hair down ...
    I am doing some mechanics questions at the moment as I am very happy!!
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    (Original post by TeeEm)
    ...
    I am nearing a solution, but you will hate me for how inelegant it is.
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    (Original post by TeeEm)
    I have let my hair down ...
    I am doing some mechanics questions at the moment as I am very happy!!
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    This image makes me very happy, thank you ^^
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    (Original post by Student403)
    :emo: I will crack down on the two once I have finished them by Easter
    Zacken is also hiding ...
    he suddenly remembered he had to shampoo his beard ....
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    (Original post by TeeEm)
    Zacken is also hiding ...
    he suddenly remembered he had to shampoo his beard ....
    (Original post by Zacken)
    I am nearing a solution, but you will hate me for how inelegant it is.
    Oi!
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    (Original post by Zacken)
    I am nearing a solution, but you will hate me for how inelegant it is.
    ready to be impressed
    last years under 18 mathmo would have gone to Gome44


    this year still a while to go, you have to earn this year's title against, a strong crowd such as physicsmaths Krollo 16Characters.... Renzhi10122 B_9710 Duke Glacia Louisb19 to name a few
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    (Original post by TeeEm)
    could you point me in the right direction?
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    Please don't kill me, I know it's inelegant!

    Your ellipse is nothing but:

    \displaystyle 

\begin{equation*}\frac{\left(x - \frac{3}{2}\right)^2}{(\frac{5}{  2})^2} + \frac{y^2}{2^2} = 1\end{equation*}

    We can shift things along using u = x - \frac{3}{2} so that our foci are at O(-3/2, 0) and T(0, 3/2).

    Now, we know that: PO = ePM' and PT = ePM, but from a simple sketch, we can see:

    PM' = u + \frac{5/2}{e} and PM = \frac{\frac{5}{2}}{e} - u.

    So: |OP| + |PT| = e\left(\frac{5/2}{e} - u\right) + e\left(\frac{5/2}{e} + u\right) = 2\times \frac{5}{2} = 5
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    (Original post by Zacken)
    Please don't kill me, I know it's inelegant!

    Your ellipse is nothing but:

    \displaystyle 

\begin{equation*}\frac{\left(x - \frac{3}{2}\right)^2}{(\frac{5}{  2})^2} + \frac{y^2}{2^2} = 1\end{equation*}

    We can shift things along using u = x - \frac{3}{2} so that our foci are at O(-3/2, 0) and T(0, 3/2).

    Now, we know that: PO = ePM' and PT = ePM, but from a simple sketch, we can see:

    PM' = u + \frac{5/2}{e} and PM = \frac{\frac{5}{2}}{e} - u.

    So: |OP| + |PT| = e\left(\frac{5/2}{e} - u\right) + e\left(\frac{5/2}{e} + u\right) = 2\times \frac{5}{2} = 2
    I think the method is fine but I think there is a number error somewhere
    (you do not post you workings instead you show off you LaTex, so I cannot locate it)
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    (Original post by TeeEm)
    I think the method is fine but I think there is a number error somewhere
    (you do not post you workings instead you show off you LaTex, so I cannot locate it)
    Is the form my ellipse correct?
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    (Original post by Zacken)
    Is the form my ellipse correct?
    I do not think so
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    (Original post by TeeEm)
    I do not think so
    I did: \displaystyle 5r - 3x = 8 \Rightarrow 25r^2 = (8+3x)^2 \Rightarrow 25(x^2 + y^2) = 64 + 48x + 9x^2, so re-arranging:

    \displaystyle 16x^2 - 48x + 25y^2 = 64 \Rightarrow 16((x-3/2)^2 - 9/4) + 25y^2 = 64

    So \displaystyle 16(x-3/2)^2 - 36 + 25y^2 = 64 Hence:

    \displaystyle 16(x-3/2)^2 + 25y^2 = 100 \Rightarrow \frac{16(x-3/2)^2}{100} + \frac{y^2}{4} = 1
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    (Original post by Zacken)
    I did: \displaystyle 5r - 3x = 8 \Rightarrow 25r^2 = (8+3x)^2 \Rightarrow 25(x^2 + y^2) = 64 + 48x + 9x^2, so re-arranging:

    \displaystyle 16x^2 - 48x + 25y^2 = 64 \Rightarrow 16((x-3/2)^2 - 9/4) + 25y^2 = 64

    So \displaystyle 16(x-3/2)^2 - 36 + 25y^2 = 64 Hence:

    \displaystyle 16(x-3/2)^2 + 25y^2 = 100 \Rightarrow \frac{16(x-3/2)^2}{100} + \frac{y^2}{4} = 1
    correct now
 
 
 
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