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Math Help - The Parabola in Parametric Form... Proving parallel lines!

  1. #1
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    Question The Parabola in Parametric Form... Proving parallel lines!

    The straight line drawn from a point P on a parabola to the vertex cuts the directrix at T. Prove that if the focus is at S, that TS is parallel to the tangent at P.

    S = (2as, as^2)
    P = (2ap, ap^2)
    T = (2at, at^2)

    gradient PT = (at^2 - ap^2)/2at - 2ap
    = a (t^2 - p^2)/ 2a(t-p)
    = (t + p)/2

    gradient TS = (at^2 - as^2)/(2at - 2as)
    = (t + s)/2

    I can't seem to get gradient PT = gradient TS

    Help please

    Thanks in advance
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  2. #2
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    Hi

    The equation you have chosen for the parabola is y = \frac{x^2}{4a}
    P(2ap,ap^2) is on the parabola.
    But S and T are not on the parabola.

    S is the focus so S(0,a)
    T is on the directrix so T(x,-a)

    You have to find T abscissa to prove that (TS) is parallel to the tangent at P.
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    Quote Originally Posted by running-gag View Post
    Hi

    The equation you have chosen for the parabola is y = \frac{x^2}{4a}
    P(2ap,ap^2) is on the parabola.
    But S and T are not on the parabola.

    S is the focus so S(0,a)
    T is on the directrix so T(x,-a)

    You have to find T abscissa to prove that (TS) is parallel to the tangent at P.
    ahhh I see.. but how would I find T abscissa?
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  4. #4
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    Quote Originally Posted by differentiate View Post
    ahhh I see.. but how would I find T abscissa?
    Through its definition : T is the intersection point between :
    - the line passing through P(2ap,apē) and the vertex of the parabola (0,0)
    - the directrix of the parabola whose equation is y=-a
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    Quote Originally Posted by running-gag View Post
    Through its definition : T is the intersection point between :
    - the line passing through P(2ap,apē) and the vertex of the parabola (0,0)
    - the directrix of the parabola whose equation is y=-a
    so I would have to solve simultaneously the equation of point P to the vertex, and y = -a

    equation Point P to the vertex
    ap^2 - 0 y - 0
    ------- = ..........
    2ap - 0 x - 0

    y2ap = xap^2

    y = 2ap/xap^2

    y = -a

    -a = 2ap/xap^2
    -xa^2p^2 = 2ap
    -xap = 2

    now what???

    soz.. I'm new to this kind of stuff
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  6. #6
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    Quote Originally Posted by differentiate View Post
    so I would have to solve simultaneously the equation of point P to the vertex, and y = -a

    equation Point P to the vertex
    ap^2 - 0 y - 0
    ------- = ..........
    2ap - 0 x - 0

    y2ap = xap^2

    ...
    2apy = ap^2x . Divide through by 2ap:

    y = \dfrac{ap^2}{2ap} x = \dfrac p2 x

    Now y = -a. Plug in this term and solve for x:

    -a=\dfrac p2 x~\implies~x = -\dfrac{2a}p

    Thus T\left(-\dfrac{2a}p\ ,\ -a\right)
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    Quote Originally Posted by earboth View Post
    2apy = ap^2x . Divide through by 2ap:

    y = \dfrac{ap^2}{2ap} x = \dfrac p2 x

    Now y = -a. Plug in this term and solve for x:

    -a=\dfrac p2 x~\implies~x = -\dfrac{2a}p

    Thus T\left(-\dfrac{2a}p\ ,\ -a\right)
    ahhh thanks... okay, after that I would need to calculate the gradient of TS and tangent at P....

    TS =
    s + a
    ----------
    0 + 2a/p

    = ps + ap
    -------------
    2a

    but what does it mean by tangent at P...would that mean PT? but that wouldn't make sense, because they wouldn't be parallel if it was PT and ST...

    soz again. I'm so stupid
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  8. #8
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    Quote Originally Posted by differentiate View Post
    ahhh thanks...
    ...
    but what does it mean by tangent at P...would that mean PT? but that wouldn't make sense, because they wouldn't be parallel if it was PT and ST...
    I assume that you have made a sketch. (see attachment)

    As you can see the line ST is parallel to the tangent in P at the parabola.

    You already have: P(2ap, ap^2) , S(0, a), T\left(-\dfrac{2a}p\ ,\ -a\right)

    The slope of the tangent in P is: m_t=\dfrac1{2a} \cdot 2ap= p Use the first derivation to calculate the slope of the tangent.

    The slope of the line ST is.

    m_{ST}=\dfrac{a-(-a)}{\frac{2a}p - 0} = p

    Thus both lines must be parallel.
    Attached Thumbnails Attached Thumbnails The Parabola in Parametric Form... Proving parallel lines!-parabtang_parallel.png  
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