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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]

(x+y)^2
(y-x) = ------- - (x+y) + 4
2*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
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations

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