FP3 geometry

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  1. snow leopard's Avatar
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    FP3 geometry
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    How do you do 8(d) ?

    The answer to 8(b) if you need it is (-ct^-3, -ct^3)
  2. snow leopard's Avatar
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    Re: FP3 geometry
    Bump...
  3. jack.hadamard's Avatar
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    Re: FP3 geometry
    By this point, you would have obtained parametric equations for the mid-point.


    So... I would start with

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    4xy = \cdots = c^2(1 - t^{4} - t^{-4} + 1) = \cdots

    and then I'd use the previous part.
  4. snow leopard's Avatar
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    Re: FP3 geometry
    (Original post by jack.hadamard)
    By this point, you would have obtained parametric equations for the mid-point.


    So... I would start with

    Spoiler:
    Show

    4xy = \cdots = c^2(1 - t^{4} - t^{-4} + 1) = \cdots

    and then I'd use the previous part.
    From what I understand, your method is to plug in x and y into the LHS and show that the expression equals the RHS i.e. zero. In the mark scheme they created the equation from scratch though, so I'm unsure if I'm permitted to do this?
  5. jack.hadamard's Avatar
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    Re: FP3 geometry
    (Original post by snow leopard)
    From what I understand, your method is to plug in x and y into the LHS and show that the expression equals the RHS i.e. zero. In the mark scheme they created the equation from scratch though, so I'm unsure if I'm permitted to do this?
    You are trying to eliminate the parameter t, and to produce a Cartesian equation. The expression will not be zero.
    Whether you will solve for the parameter and substitute it in, or just observe that xy has a nice expression, it should not matter.
  6. snow leopard's Avatar
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    Re: FP3 geometry
    (Original post by jack.hadamard)
    You are trying to eliminate the parameter t, and to produce a Cartesian equation. The expression will not be zero.
    Whether you will solve for the parameter and substitute it in, or just observe that xy has a nice expression, it should not matter.
    I see, I needed to eliminate t. It makes sense now thanks.

    The only thing left troubling me is how (t^4 + t^-4 - 2) factorises to (t^-2-t^2)^2... do you just notice that by inspection?
  7. jack.hadamard's Avatar
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    Re: FP3 geometry
    (Original post by snow leopard)
    The only thing left troubling me is how (t^4 + t^-4 - 2) factorises to (t^-2-t^2)^2... do you just notice that by inspection?
    If you have done enough hyperbolic functions, you'd notice these quickly; e.g. 4\sinh^2(2x) = e^{4x} - 2 + e^{-4x}.

    Otherwise, you could approach it by:

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    \displaystyle t^4 + t^{-4} - 2 = t^4 + \frac{1}{t^4} - 2 = \frac{1}{t^4}(t^8 - 2t^4 + 1) = \frac{1}{t^4}(t^4 - 1)^2 = (t^2 - t^{-2})^2 = (t^{-2} - t^2)^2.

    However, it is better to use the questions to guide you.. you see a square, you should be getting a square.
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