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# FP3 geometry Tweet

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1. 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. Re: FP3 geometry
Bump...
3. Re: FP3 geometry
By this point, you would have obtained parametric equations for the mid-point.

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and then I'd use the previous part.
4. Re: FP3 geometry
By this point, you would have obtained parametric equations for the mid-point.

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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. 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 , 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 has a nice expression, it should not matter.
6. Re: FP3 geometry
You are trying to eliminate the parameter , 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 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. 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. .

Otherwise, you could approach it by:

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.

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