FP1 - Parabola question
Maths and statistics discussion, revision, exam and homework help.
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Re: FP1 - Parabola questionThe 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.(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 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.