We will consider a
parabol and trace out its pedal curve using Jeometry.
First we need to construct a parabola. This is very easy
if we recall that it is the locus of a point T whose
distance from a fixed point F is equal to the distance
from a fixed line d. So we
pick up a point D along d and
build the perpendicular from D to d.
This line will meet the axis of DF at T.
Now, draw a point O on the plane and build the
perperdicular from O to the tangent MT: its foot S traces
out the pedal as T moves (see the picture below).
We can get a parametrization of our pedal in this way.
Without loss of generality, we can suppose that the
parabola has the Cartesian equation y = kx2
and O has coordinates (a,
b). A
generic point T of the parabola will have coordinates (t, kt2)
and the tangent at T to the parabola will be y-kt2=2kt
(x - t). The perpendicular from O to this
tangent has equation -2kt(y
- b) = x - a. Solving for x and y
| y - kt2
= 2kt (x - t) |
(1) |
| -2kt(y
- b) = x - a |
You can use the Groebner
applet to solve the system
(even if you can easily do it by yourself).
To eliminate y, find the 1st elimination
ideal <f1>
of <y-kt2-2kt(x-t), 2kt(y-b)+x-a>
by computing the Groebner basis w.r.t. the
1st-elimination
order with y>x>t>k>b>a,
then solve f1=0 for x.
To eliminate x, find the 1st elimination
ideal <g1>
of <y-kt2-2kt(x-t), 2kt(y-b)+x-a>
by computing the Groebner basis w.r.t. the
1st-elimination
order with x>y>t>k>b>a,
then solve g1=0 for y.
See Solving systems of polynomial
equations for more
details. |
|
we get
| |
2k2t3+2bkt+a |
(2) |
| x= |
------------------ |
| |
1+4k2t2 |
| |
|
| |
(4bk2-k)t2+2akt |
| y= |
--------------------. |
| |
1+4k2t2 |
To find the Cartesian equation of the pedal we could
apply the rational
implicitization procedure to (2),
but it is much easier to work it out directly from (1). With the Groebner applet we find the
1st elimination ideal of I = <y-kt2-2kt(x-t),
2kt(y-b)+x-a> by computing the Groebner
basis w.r.t. to lex order with t>x>y>k>b>a
(we must treat k, b, a
as variables since we can't use parametric coefficients
with the applet). We get
I1
= <kx2y -
bkx2 + 1/4x2
- akxy
+ abkx - 1/2ax
+ ky3 - 2bky2
+ b2ky + 1/4a2>.
Is every point of V(I1)
on (1)? This is the same as asking if
every partial solution (x,y) of (1) extends to a complete solution (x, y, t). Over C
we can give a positive answer using the Extension Theorem.
And over R? Looking at the generators in the
Groebner basis
g1
= kx2y - bkx2
+ 1/4x2
- akxy
+ abkx - 1/2ax
+ ky3 - 2bky2
+ b2ky + 1/4a2
g2
= k(y - b)t + 1/2x
- 1/2a
g3
= (x - a)t - 2x2
+ 2ax -2y2 + 2by
g4
= kt2 - 2akt -
4kx2 + 4akx - 4ky2
+ 4bky + y
we see that t is real and uniquely
determined whenever (x, y) <>
(a, b) (we use g2 and g3). When (x, y) = (a, b), it follows
from g4
= 0 that t is real if and only if D
= a2k2
- bk >= 0. Hence, if D < 0, V(I1)introduces one point (a,
b)which is not on (1):
the screenshot below shows the pedal in one of these
cases.
 |
| V(I1) covers the isolated point O
whereas O is not on the pedal |
The pedal will pass through (0, 0) if we pick O on the
y-axis (i.e. on the axis of the parabola). So, letting a = 0we get
| kx2y
- bkx2 + 1/4x2
+ ky3 - 2bky2
+ b2ky = 0 |
(3). |
Furthermore, we have four special cases depending on
the particular position of O along the axis.
- O is the intersection of the axis and the
directrix. The curve is also known as right
strophoid (more generally, if O is on the
directrix but not on the axis of the parabola, we
have an oblique strophoid). We have
used Jeometry to redraw the pedal in this special
case.
The Cartesian equation can be obtained by
letting b = -1/4k;
substituting into (3) we have
kx2y
+ 1/2x2
+ ky3 + 1/2y2
+ 1/16ky
= 0.
With the translation
we get
| x2(4ky
+ 1) + y2(4ky
- 1) = 0 |
(4). |
- O is the reflection of the focus on the directrix
of the parabola. The pedal is called the trisectrix
of Maclaurin. We can draw it using
Jeometry:
The Cartesian equation can be obtained by
letting b = -3/4k;
substituting into (3) we get
kx2y
- 1/2x2
+ ky3 + 3/2y2
+ 9/16ky
= 0 .
With the translation
we have
| x2(4ky
+ 1) + y2(4ky
- 3) = 0 |
(5). |
- O is the vertex of
the parabola. The curve is also known as the cissoid of
Diocles.
The Cartesian equation can be obtained by
letting b = 0; substituting into
(3) we get
| kx2y
+ 1/4x2
+ ky3
= 0 |
(6). |
- O is the focus of the parabola. The pedal is
simply a line (the line traced out by M).
|