Associated Topics || Dr. Math Home || Search Dr. Math

### Points Where a Parametric Equation Crosses Itself

```
Date: 10/12/2001 at 12:59:21
From: Jessica Stover
Subject: Find all points where a parametric equation crosses itself

I am supposed to find all the points where the function crosses
itself of the parametric equation: x=t(9-t^2) and y=(t^2-1)(t^2-9) on
the interval -4<t<4. I don't even know where to start.  Could you

Thank you,
Jessica Stover
```

```
Date: 10/12/2001 at 16:04:44
From: Doctor Rob
Subject: Re: Find all points where a parametric equation crosses
itself

Thanks for writing to Ask Dr. Math, Jessica.

One way to proceed is to call the two times that have the same x and
y values t and T.  Then you have the equations

t*(9-t^2) = T*(9-T^2),
(t^2-1)*(t^2-9) = (T^2-1)*(T^2-9).

These you need to solve for t and T, subject to the condition that
t - T is nonzero.

One way to solve these equations is to eliminate one of the variables,
say t, from the two equations, to give you one equation in the unknown
T, and then solve that. The equation you get has degree 6, and factors
into two linear factors times a biquadratic factor. This allows you to
solve for the roots explicitly. There seem to be 6 real roots, all in
the indicated interval. Once you know the values of T, you can find
the values of x and y.

Good luck!

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/15/2001 at 20:04:09
From: jstover
Subject: Re: Find all points where a parametric equation crosses
itself

Dr. Math,

Thank you for all of your help, but I was wondering if you could go
through the steps for solving for T. I cannot figure it out or come up
with 6 answers when I do. Thank you again.

Jessica Stover
```

```
Date: 10/16/2001 at 10:00:56
From: Doctor Rob
Subject: Re: Find all points where a parametric equation crosses
itself

Thanks for writing back, Jessica.

Expand both equations

(1)  -t^3 + 9*t = -T^3 + 9*T,
(2)  t^4 - 10*t^2 + 9 = T^4 - 10*T^2 + 9,

Add t times the (1) to (2):

-t^2 + 9 = T^4 - t*T^3 - 10*T^2 + 9*t*T + 9.
(3)  -t^2 = T^4 - t*T^3 - 10*T^2 + 9*t*T.

Now add -t times (3) to (1):

(4)  9*t = -t*T^4 + (t^2-1)*T^3 + 10*t*T^2 - 9*(t^2-1)*T.

Solve (3) for t^2:

(5)  t^2 = -T^4 + t*T^3 + 10*T^2 - 9*t*T.

Substitute (5) in (4):

0 = -t*T^4 + ([-T^4 + t*T^3 + 10*T^2 - 9*t*T]-1)*T^3 + 10*t*T^2 -
9*([-T^4 + t*T^3 + 10*T^2 - 9*t*T]-1)*T - 9*t,
(6)  0 = t*(T^6-19*T^4+91*T^2-9) + (-T^7-19T^5+91*T^3-9*T).

Factor this into

(7)  0 = (t-T)*(T^6-19*T^4+91*T^2-9).

Since t - T is nonzero, that factor can be dropped.  Thus

(8)  0 = T^6 - 19*T^4 + 91*T^2 - 9.

This is the 6th degree equation for T.  It can be further factored
as follows:

0 = (T^2 - 9)*(T^4 - 10*T^2 + 1),
(9)  0 = (T - 3)*(T + 3)*(T^4 - 10*T^2 + 1).

The first two factors give

(10)  T = 3,
(11)  T = -3.

The last factor gives

(12)  T^2 = 5 + 2*sqrt(6),
(13)  T^2 = 5 - 2*sqrt(6),

which in turn yield

(14)  T = sqrt(5+2*sqrt[6]),
(15)  T = -sqrt(5+2*sqrt[6]),
(16)  T = sqrt(5-2*sqrt[6]),
(17)  T = -sqrt(5-2*sqrt[6]).

Thus (10), (11), and (14)-(17) are the six possible values of T.

I leave it to you to find the corresponding values of t (which
must be chosen from the same list of 6 values), subject to t - T
being nonzero.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search