Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5.

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Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5.

Mathematics
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Center is (0,0)
have you sketched the focus and directrix to determine which way the parabola opens...?

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It opens right
there are 2 ways to do this question... 1. use the distance formula... knowing the distance from the focus to a point on the parabola is equal to the distance from the point to the directirx. 2. find the focal length and use a standard form of the parabola.
great so the standard from I use for this parabola is \[(y - k)^2 = 4a(x - h)\] where (h, k) is the vertex and a is the focal length. any ideas on the focal length..?
I have no idea what that is.
wait. it's 5
I just remembered what that was
ok... so it seems you have only been taught about selecting a point on the parabola.... |dw:1436392195951:dw| do by definition, the distance from the focus to P is equal to the distance from P to Q I'd expect that's the way you have been taught to find the locus of the parabola.
|dw:1436392369178:dw| so which method do you want to use..?
the line of symmetry for the parabola is y = 0 the vertex is halfway between the directrix and focus.. so any ideas of the location of the vertex and the focal length..?
I dont care. Whichever you are better with or think is easier.
Vertex is (0,0) and focal length is 5
ok... so how far between the directrix and focus?
10
great so 2a = 10 so a = 5 is that ok...?
yup
the vertex in this question is to the left of the focus... and it is the focal length to the left on the line of symmetry any ideas in where the vertex is..?
0,0?
great so you know h = 0, k =0 and a = 5 plug them into the standard form \[(y - k)^2 = 4a(x - h)\] (h, k) is the vertex and a is the focal length
one sec

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