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July 5th, 2012, 03:42 AM   #1
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Finding equation of parabola with focus and directrix

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Given directrix y = -x + 2 and focus (0,0), find the equation of the parabola
So I found the equation for the perpendicular line to the directrix in order to find the vertex, which I got the line y = x that is perpendicular to the directrix, then solved the system of equations to find the common intersection point, which was (1,1). I used the midpoint formula from the focus to the directrix, and got (0.5,0.5) (even though I know by intuition). So the vertex is at (0.5,0.5) since it is exactly in between the focus and directrix.

Using the form , I plugged in the vertex already. To find the value of 'p' I needed to use the distance formula for between either the focus and vertex, or the vertex and directrix which I know should be exactly the same. So I did:




So:





Put in standard form:



I don't know if I've made any careless errors and I didn't want to rationalize the denominator and risk messing this up, but when I try to graph this, the graph of the parabola passes through the directrix...I thought it wasn't supposed to touch it at all?
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July 5th, 2012, 08:15 AM   #2
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Re: Finding equation of parabola with focus and directrix

The axis of symmetry of the parabola is perpendicular to the directrix. If the directrix is y = -x + 2, your parabola should be oriented at a 45 degree angle.
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July 5th, 2012, 08:25 AM   #3
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Re: Finding equation of parabola with focus and directrix

When you write
"Using the form...", you are using the form of a VERTICAL parabola.

Your equation will:
1. Contain an "xy" term, and
2. Not be a function of x (I.e. not pass the vertical line test)
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July 5th, 2012, 09:44 AM   #4
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Re: Finding equation of parabola with focus and directrix

One method we could use is to find the locus of all points equidistant from the focus and directrix:







Here is a plot of the parabola, directrix, and the axis of symmetry for the parabola:

[attachment=0:1qw49g56]slantparabola.jpg[/attachment:1qw49g56]
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File Type: jpg slantparabola.jpg (24.2 KB, 325 views)
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