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# fp1 focus,directrix and parabolas

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1. http://files.physicsandmathstutor.co...%20Edexcel.pdf

question 2 here

I know that since the co-ordinates which will be equidistant away from the point and line. It'll form a parabola which is a U shape with the curved lines pointing left.

But how am i supposed to work out the equation of that curved line?

is it just y=-4ax? or do i need to show some proof of that?
2. You can use you wouldn't have to show proof.
3. (Original post by NotNotBatman)
You can use you wouldn't have to show proof.
so y^2 =4ax

and since it's on the other half of the graph the it's a minus and since you got a point -3 a=-3

thanks
4. tfw on question 5 when u did everything right and then u did z=x^-1 y instead of z=yx^-1 on the last part

5. It always helps to be able to derive these in an emergency

take a generic point (x,y)
distance to point (-3,0) is sqrt((x+3)^2 + y^2)
distance to line x = 3 is x-3

these must be equal

x-3 = sqrt((x+3)^2 + y^2)
(x-3)^2 = (x+3)^2 + y^2
x^2 - 6x + 9 = x^2 + 6x + 9 + y^2
y^2 = -12x
x = -y^2/12

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