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

Quote:
 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 $(x - \frac{1}{2})^{2}= 4p(y - \frac{1}{2})$, 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: $distance= \sqrt{(\frac{1}{2} - 0)^{2} + (\frac{1}{2} - 0)^{2}}$
$distance= \sqrt{(\frac{1}{4}) + (\frac{1}{4})}$
$distance= \sqrt{\frac{2}{4}}$
$distance= \frac{\sqrt{2}}{2}$

So:

$(x - \frac{1}{2})^{2}= 4(\frac{\sqrt{2}}{2})(y - \frac{1}{2})$
$(x - \frac{1}{2})^{2}= \frac{4\sqrt{2}}{2}(y - \frac{1}{2})$
$(x - \frac{1}{2})^{2}= 2\sqrt{2}(y - \frac{1}{2})$

Put in standard form:

$y= \frac{x^{2} - x + \frac{4\sqrt{2} + 1}{4}}{2\sqrt{2}}$

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?

 July 5th, 2012, 08:15 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1134 Math Focus: Elementary mathematics and beyond 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.
 July 5th, 2012, 08:25 AM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 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)
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:

$\sqrt{x^2+y^2}=\frac{\|x+y-2\|}{\sqrt{2}}$

$2x^2+2y^2=x^2+y^2+2xy-4x-4y+4$

$x^2+y^2-2xy+4x+4y-4=0$

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

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

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