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Finding equation of parabola with focus and directrix

daigo

Member
Jun 27, 2012
60
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 [tex](x - \frac{1}{2})^{2} = 4p(y - \frac{1}{2})[/tex], 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: [tex]distance = \sqrt{(\frac{1}{2} - 0)^{2} + (\frac{1}{2} - 0)^{2}}[/tex]
[tex]distance = \sqrt{(\frac{1}{4}) + (\frac{1}{4})}[/tex]
[tex]distance = \sqrt{\frac{2}{4}}[/tex]
[tex]distance = \frac{\sqrt{2}}{2}[/tex]

So:

[tex](x - \frac{1}{2})^{2} = 4(\frac{\sqrt{2}}{2})(y - \frac{1}{2})[/tex]
[tex](x - \frac{1}{2})^{2} = \frac{4\sqrt{2}}{2}(y - \frac{1}{2})[/tex]
[tex](x - \frac{1}{2})^{2} = 2\sqrt{2}(y - \frac{1}{2})[/tex]

Put in standard form:

[tex]y = \frac{x^{2} - x + \frac{4\sqrt{2} + 1}{4}}{2\sqrt{2}}[/tex]

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?
 

soroban

Well-known member
Feb 2, 2012
409
Hello, daigo!

given directrix .[tex]y \:=\:\text{-}x+2[/tex] and focus [tex](0,0)[/tex], find the equation of the parabola.

Sorry ... you started off wrong . . .

The directrix is a slanted line.
The parabola has been rotated.
The equation will have an [tex]xy[/tex]-term.

The graph looks like this . . .

Code:
          o |
           2o
            | o
            |   o       .
            |     o   .
            |       o
      *     *     .   o
            |   *       o
            | .           o
      ------o------*--------o----
            |F              2 o
        .   |                   o
      .     |      *
            |
 

daigo

Member
Jun 27, 2012
60
So I should not use the (x-h)^2 = 4p(y - k) form and there's a different equation to be used?

By the way, how did you create that ASCII graph? Surely it'd take too much of your time to draw it by hand
 

soroban

Well-known member
Feb 2, 2012
409
Hello, daigo!

By the way, how did you create that ASCII graph?
Surely it'd take too much of your time to draw it by hand.

I created a procedure several years ago.
Since then I've explained it on a few other math-sites,
. . but never on this site.
Besides, that was a few years ago
. . and my mini-lessons are surely buried.

It is quite a bit of work, but I enjoy it:
. . the challenge, the planning, etc.


Obviously, I go into CODE mode.

First, I type this line: period-period-period-hyphen-hyphen-hyphen . . .
. . with a space between them.
Code:
. . . - - - - - - - - - - - -
Then I use COPY/PASTE to make, say, four such lines.
Code:
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - - 
. . . - - - - - - - - - - - - 
. . . - - - - - - - - - - - -
Then I use COPY/PASTE again to make a few more such sets.
Code:
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - - 
. . . - - - - - - - - - - - - 
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
. . . - - - - - - - - - - - -
Then I carefully insert other symbols,
. . replacing the hyphens.
Code:
. . . - - | - - - - - - - - -
. . . - * | - - -*- - - - - - 
. . . - - * - - * - - - - - - 
. . . - - | * -*- - - - - - -
. . . - - | - * - - - - - - - 
. . . - - | -*- * - - - - - -
. . . - - + * - - * - - - - - 
. . . - - |*- - - - * - - - -
. . . - - * - - - - - * - - -
. . . - -*| - - - - - - * - -
. . . - * | - - - - - - - * -
. . . - - | - - - - - - - - - -
Then I eliminate the unwanted symbols.
At the far right, I simply delete the hyphens.
At the left, I replace each hyphen with a space.
Code:
          |
        * |      * 
          *     * 
          | *  *
          |   * 
          |  *  *
      ----+-*-----*---------- 
          |*        *
          *           *
         *|             *
        * |               *
          |

One of my favorites is a large circle.
Code:
. . . - - - - * * * - - - - -
. . . - - * - - - - - * - - - 
. . . - * - - - - - - - * - - 
. . . -*- - - - - - - - -*- -
. . . - - - - - - - - - - - -
. . . * - - - - - - - - - * -
. . . * - - - - * - - - - * -
. . . * - - - - - - - - - * -
. . . - - - - - - - - - - - -
. . . -*- - - - - - - - -*- -
. . . - * - - - - - - - * - -
. . . - - * - - - - - * - - -
. . . - - - - * * * - - - - -
I like to leave a margin at the left of the diagram.
That's the reason for the three periods.
 

daigo

Member
Jun 27, 2012
60
Wow, you really do draw it manually! Surely there must be a way to code a small script where you input a formula and it draws out a graph maybe. Anyway, that's cool
 

earboth

Active member
Jan 30, 2012
74
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 [tex](x - \frac{1}{2})^{2} = 4p(y - \frac{1}{2})[/tex], 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: [tex]distance = \sqrt{(\frac{1}{2} - 0)^{2} + (\frac{1}{2} - 0)^{2}}[/tex]
[tex]distance = \sqrt{(\frac{1}{4}) + (\frac{1}{4})}[/tex]
[tex]distance = \sqrt{\frac{2}{4}}[/tex]
[tex]distance = \frac{\sqrt{2}}{2}[/tex]

So:

[tex](x - \frac{1}{2})^{2} = 4(\frac{\sqrt{2}}{2})(y - \frac{1}{2})[/tex]
[tex](x - \frac{1}{2})^{2} = \frac{4\sqrt{2}}{2}(y - \frac{1}{2})[/tex]
[tex](x - \frac{1}{2})^{2} = 2\sqrt{2}(y - \frac{1}{2})[/tex]

Put in standard form:

[tex]y = \frac{x^{2} - x + \frac{4\sqrt{2} + 1}{4}}{2\sqrt{2}}[/tex]

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?
1. Use the definition of the parabola as the locus of points whose distance from a straight line and a fixed point are equal.

2. Let P(x ,y) denotes the point on the parabola. Then the distance from the focus (= origin) is

[TEX]\displaystyle{d_1=\sqrt{x^2+y^2}}[/TEX]

3. The distance of a point Q(m, n) from the line Ax + By + C= 0 is

[TEX]\displaystyle{D=\frac{Am+Bn+C}{\sqrt{A^2+B^2}}}[/TEX]

In your case the line has the equation: [TEX]\displaystyle{x+y-2=0}[/TEX]

and A = 1 and B = 1 and C = -2

4. Now screw those bits and pieces together:

[TEX]\displaystyle{\sqrt{x^2+y^2}=\frac{x+y-2}{\sqrt{1+1}}}[/TEX]

5. This equation describes the parabola. After a few steps you should come out with:

[TEX]\displaystyle{(x+2)^2+(y+2)^2-2xy-12=0}[/TEX]