Finding equation of parabola with focus and directrix

  • Context: MHB 
  • Thread starter Thread starter hatelove
  • Start date Start date
  • Tags Tags
    Focus Parabola
Click For Summary

Discussion Overview

The discussion revolves around finding the equation of a parabola given a focus at (0,0) and a directrix defined by the line y = -x + 2. Participants explore the geometric properties of the parabola, including the vertex's location and the implications of the directrix on the parabola's shape and equation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes their method for finding the vertex by determining the perpendicular line to the directrix and solving for the intersection point, arriving at (0.5, 0.5) as the vertex.
  • The same participant uses the standard form of a parabola to derive the equation, expressing concern about the graph intersecting the directrix, which they believed should not occur.
  • Another participant points out that the directrix is slanted and suggests that the parabola has been rotated, implying that a different equation involving an xy-term may be necessary.
  • There is a question raised about the appropriate form of the parabola equation to use, indicating uncertainty about the initial approach taken by the first participant.
  • One participant shares their method for creating an ASCII graph, indicating it is a labor-intensive process that they enjoy, while another expresses curiosity about the possibility of automating this process with a script.
  • A later reply provides a step-by-step breakdown of the definition of a parabola and suggests a different equation based on the distances from the focus and the directrix, leading to a new equation that participants can explore.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct form of the parabola's equation. There are competing views regarding the implications of the directrix and the appropriate mathematical approach to take.

Contextual Notes

Participants express uncertainty about the correctness of their calculations and the implications of their findings, particularly regarding the relationship between the parabola and the directrix. There are also unresolved questions about the proper form of the parabola's equation given the slanted directrix.

hatelove
Messages
101
Reaction score
1
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?
 
Mathematics news on Phys.org
Hello, daigo!

given directrix .y \:=\:\text{-}x+2 and focus (0,0), 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 xy-term.

The graph looks like this . . .

Code: 
          o |
           2o
            | o
            |   o       .
            |     o   .
            |       o
      *     *     .   o
            |   *       o
            | .           o
      ------o------*--------o----
            |F              2 o
        .   |                   o
      .     |      *
            |
 
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
 
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.
 
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
 
daigo said:
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?

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

\displaystyle{d_1=\sqrt{x^2+y^2}}

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

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

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

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

4. Now screw those bits and pieces together:

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

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

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

Similar threads

MHB Find vertex, focus, and directrix of parabola: y^2+12y+16x+68=0
  • · Replies 7 ·
Replies
7
Views
3K
High School Parabola vs Hyperbola, why does a Hyperbola have two foci/curves?
  • · Replies 3 ·
Replies
3
Views
2K
MHB Find Equation of Parabola Given Focus & Directrix - Jose's Q&A
  • · Replies 1 ·
Replies
1
Views
2K
MHB Finding focus, given four tangents
  • · Replies 6 ·
Replies
6
Views
3K
MHB Parabola Tangent: GP Relation for Fixed Point Chords
  • · Replies 4 ·
Replies
4
Views
2K
MHB Equation of ellipse from directrix and focus
  • · Replies 1 ·
Replies
1
Views
7K
MHB Finding the equation of parabola
  • · Replies 2 ·
Replies
2
Views
2K
High School Complete the square, yes, but what does it mean?
  • · Replies 19 ·
Replies
19
Views
3K
MHB Fixed point,, Jacobi- & Newton Method, Linear Systems
  • · Replies 7 ·
Replies
7
Views
2K
Graduate What Are Conic Sections and How Are They Defined?
  • · Replies 1 ·
Replies
1
Views
4K