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    I've previously learnt that this is the general equation of a parabola :
    1/2(b-k) * (x-a)^2 = y-((b+k)/2)

    But my current book says that it is represented by
    y^2=4ax

    I was wondering if someone could confirm/show that they are the same thing...
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    Well any graph of the form y=ax^2+bx+c or x=ay^2+by+c is a parabola, so both your equations give parabolae.

    However we can consider parabolae that go through the origin. If the directrix is the line x=-a and the focus is the point (a,0), then at any point on the graph, the horizontal distance from the directrix (i.e. x-(-a)) must be equal to the distance from the focus (i.e. \sqrt{(x-a)^2+y^2}). That is:

    x+a=\sqrt{(x-a)^2+y^2}

    Squaring both sides and rearranging then gives y^2=4ax.

    Then we can drop the requirement that the parabola go through the origin by translating, meaning that a general parabola is given by (y-p)^2 = 4a(x-q), which can be rearranged into the form x=ay^2+by+c. We can then switch x and y if we need to.
 
 
 
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Updated: March 22, 2011
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