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# Parabola help watch

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1. I don't want answers, but would be grateful if you could explain how I go about solving this type of problem, as we've only done ones were the parabola intersects the origin so far.

Find, in cartesian form, an equation of the parabola whose focus and directrix are respectively,
(3,0); x+5=0

and how does it change when its a X^2, rather than Y^2, for example

(0,4) ; y+4=0

Thanks alot
2. Ok, the focus and directrix are equi-distant from the vertex of a parabola. The focus is inside the concave part of the parabola, the directrix is a line perpendicular to the axis of the parabola. That means that the vertex of your parabola will be halfway between the point (3,0) and the line x=-5. Without any information that allows you to find the shape of the parabola, I guess assume it's a generic (unmodified) parabola.

As for vs. , one opens up, one opens to the right.
3. (Original post by D-Day)
Ok, the focus and directrix are equi-distant from the vertex of a parabola. The focus is inside the concave part of the parabola, the directrix is a line perpendicular to the axis of the parabola. That means that the vertex of your parabola will be halfway between the point (3,0) and the line x=-5. Without any information that allows you to find the shape of the parabola, I guess assume it's a generic (unmodified) parabola.

As for vs. , one opens up, one opens to the right.
Thanks . How do I translate that into an actual formula though, because I agree with everything you've just said, but can't think of a way of answering it still.
4. What point is halfway between (3,0) and x=-5?
5. (-1,0). But, that doesn't give the equation of the formula. I know its shifted 'one to the left'- which would is f(x+1), but i don't understand how this transforms the formula.

If the the cartesian equation for the parabola is y^2=4ax, then what does the f(x+1) transformation do to it basically?

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Updated: November 16, 2008
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