I need help with Precalc on parabolas
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# I need help with Precalc on parabolas

[From: ] [author: ] [Date: 11-05-19] [Hit: ]
where p is the focus, so the equation is y^2=4(1/2)x, which is y^2=2x.-x = -1/2 is a VERTICAL line, so the parabola is HORIZONTAL.The vertex lies to the right of the directrix,......
Find the equation of the following parabola.
Directrix is the line x=-1/2
Vertex (0,0)

How do I do this?

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If the directrix is the line x=-1/2, then you know that the focus is (1/2, 0). Since the focus is on the x-axis and is positive, you know that the equation is y^2=(something)x. The standard form for an equation of a parabola with vertex (0,0) is y^2=4px, where p is the focus, so the equation is y^2=4(1/2)x, which is y^2=2x.

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x = -1/2 is a VERTICAL line, so the parabola is HORIZONTAL.
The vertex lies to the right of the directrix, so the parabola opens to the right.

equation of a right-opening parabola:
x = a(y−k)² + h
a > 0
vertex is (h, k)
p = 1/(4a)
focus (h+p, k)
directrix x = h−p