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Finding Equations of Parabolas

Date: 12/12/96 at 19:02:20
From: Kris Duguay
Subject: Conics (parabolas)

The problem that I am pondering and trying to solve deals with the 
special conic, the parabola.  I have drawn the graph and plotted the 
given information and I just cannot see where to begin.  I would 
dearly appreciate it if you could help me with the following:

Determine the equation of the parabola whose vertex is at (1,3) and 
whose directrix is y=x.

Date: 12/12/96 at 21:56:00
From: Doctor Pete
Subject: Re: Conics (parabolas)

The directrix is related to the focus by the fact that the parabola is 
the locus of points for which the distance to the focus and 
perpendicular distance to the directrix is equal.  There are a few 
ways to go about solving this problem, but I think the easiest would 
be to introduce a transformed coordinate system in which the directrix 
is parallel to an axis; that is, let's introduce a rotated coordinate 
system (x', y'), in which the x'-axis is Pi/4 degrees clockwise of the 
x-axis.  To do this, recall that a transformed coordinate axis which 
is rotated about the origin through an angle T is given by:

     [ x' ]   [  Cos T   Sin T ][ x ]   [  x Cos T + y Sin T ]
     |    | = |                ||   | = |                    | 
     [ y' ]   [ -Sin T   Cos T ][ y ]   [ -x Sin T + y Cos T ]

So if T = Pi/4, Cos T = Sin T = 1/Sqrt[2], and this becomes:

       x' = (x + y)/Sqrt[2]
       y' = (y - x)/Sqrt[2]

Note that under this transformation, the line y = x becomes the line 
y' = 0, and we see that the vertex becomes:

 (1,3) --> ((1+3)/Sqrt[2]', (3-1)/Sqrt[2]') = (2*Sqrt[2]', Sqrt[2]')

Now, how do we find the equation of the parabola under this 
transformed coordinate system?  It's easy because we now know that the 
distance from the vertex to the directrix is simply Sqrt[2]'.  Thus 
the focus must be located at:

     F' = (2*Sqrt[2]', 2*Sqrt[2]')

since the vertex is halfway between the directrix and focus.  Now, we 
use the focus-directrix definition of the parabola to find its 
equation.  If some point (x', y') is on the parabola, then its 
perpendicular distance to the directrix is simply y'.  This must then 
be equal to its distance to the focus, so: 
Sqrt[(x' - 2*Sqrt[2])^2 + (y' - 2*Sqrt[2])^2] = y'  

Simplifying, we obtain:

     (x' - 2*Sqrt[2])^2 + (y' - 2*Sqrt[2])^2 = (y')^2
     (x')^2 - 4*Sqrt[2] x' + 8 + (y')^2 - 4*Sqrt[2] y' + 8 = (y')^2
     (x')^2 - 4*Sqrt[2] x' + 16 = 4*Sqrt[2] y'
     y' = (x')^2/(4*Sqrt[2]) - x' + 4/Sqrt[2]
        = (x')^2/(4*Sqrt[2]) - x' + 2*Sqrt[2] 

This is our equation for the parabola in the transformed coordinate 
system.  Thus, to get the equation of the parabola in the original 
coordinate system, we apply the transformations to the equation.  
Wherever we see an x', we replace it by (x+y)/Sqrt[2], and wherever we 
see a y', we replace it by (y-x)/Sqrt[2].  This gives us:

      (y-x)    ((x+y)/Sqrt[2])^2    (x+y)
     ------- = ----------------- - ------- + 2*Sqrt[2]
     Sqrt[2]       4*Sqrt[2]       Sqrt[2]

       (y-x) = ------- - (x+y) + 4

      8(y-x) = x^2 + 2xy + y^2 - 8(x+y) + 32

     x^2 + 2xy + y^2 - 16y + 32 = 0 

This is the desired result.  If we substitute the point (1,3) into 
this equation, it indeed returns 0.

Note that I didn't graph a thing - the entire solution is algebraic, 
although it often helps to picture what's going on, especially when 
one uses rotation matrices.  While the solution I have given seems 
quite long, it isn't all that much computation compared to trying to 
do it without rotating the coordinate axes.  The key step is to 
understand the focus-directrix relationship; the idea of the directrix 
has a direct geometric interpretation, which generalizes to any conic, 
not just the parabola.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations

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