# Finding equation of parabola with focus and directrix

• Jul 5th 2012, 03:45 AM
daigo
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 $\displaystyle (x - \frac{1}{2})^{2} = 4p(y - \frac{1}{2})$, 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: $\displaystyle distance = \sqrt{(\frac{1}{2} - 0)^{2} + (\frac{1}{2} - 0)^{2}}$
$\displaystyle distance = \sqrt{(\frac{1}{4}) + (\frac{1}{4})}$
$\displaystyle distance = \sqrt{\frac{2}{4}}$
$\displaystyle distance = \frac{\sqrt{2}}{2}$

So:

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

Put in standard form:

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

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?
• Jul 5th 2012, 05:17 AM
Reckoner
Re: Finding equation of parabola with focus and directrix
Quote:

Originally Posted by daigo
Using the form $\displaystyle (x - \frac{1}{2})^{2} = 4p(y - \frac{1}{2})$

Here's your problem: only parabolas with vertical or horizontal axes of symmetry will have that form. For a general parabola in the plane with directrix $\displaystyle ax + by + c = 0$ and focus $\displaystyle (h, k),$ the equation is

$\displaystyle \frac{(ax + by + c)^2}{a^2 + b^2} = (x - h)^2 + (y - k)^2.$

Using this form, I come up with $\displaystyle x^2 - 2xy + y^2 + 4x + 4y - 4 = 0.$
• Jul 5th 2012, 06:26 AM
daigo
Re: Finding equation of parabola with focus and directrix
That certainly makes a lot more sense...if I use that equation for every single parabola, including the ones with vertical/horizontal axes of symmetry, will I obtain the same result as if I used the other equation (in my original post)?
• Jul 5th 2012, 06:30 AM
Reckoner
Re: Finding equation of parabola with focus and directrix
Quote:

Originally Posted by daigo
That certainly makes a lot more sense...if I use that equation for every single parabola, including the ones with vertical/horizontal axes of symmetry, will I obtain the same result as if I used the other equation (in my original post)?

You should get the same result yes. It just might involve a little more work.