FP3 geometry
Maths and statistics discussion, revision, exam and homework help.
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Re: FP3 geometryFrom 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?(Original post by jack.hadamard)
By this point, you would have obtained parametric equations for the mid-point.
So... I would start with
and then I'd use the previous part. -
Re: FP3 geometryYou are trying to eliminate the parameter(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?
, 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.
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Re: FP3 geometryI see, I needed to eliminate t. It makes sense now thanks.(Original post by jack.hadamard)
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.
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? -
Re: FP3 geometryIf you have done enough hyperbolic functions, you'd notice these quickly; e.g.(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?
.
Otherwise, you could approach it by:
However, it is better to use the questions to guide you.. you see a square, you should be getting a square.
