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# FP1 - Parabola question Tweet

Maths and statistics discussion, revision, exam and homework help.

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1. Re: FP1 - Parabola question
(Original post by Aeyuin)
Sorry, could you explain this again? I understand how parametric equations work but I don't see how you've formed this one.

x = at^2 ; y = 2at ; y^2 = 4ax

Basically, where did these values come from?

And why is mc, a? I've never even seen a question where the intercept x the gradient was significant...
The first bit is standard bookwork. The parabola y^2 = 4ax (focus at (a,0), directrix x = -a) can be represented as (at^2, 2at). So I quoted this without justification.

The question defined the general tangent as y = mx +c.

The general tangent came out as y = (1/t)x + at

So m = 1/t , c = at, and hence mc = a which is what the question wants - a constant. Importantly this result has no 't' in it. This is just a property of parabolas.
2. Re: FP1 - Parabola question
What exam board is this question from. I do AQA and we never get questions like this....