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

### Using Descartes' Method of Tangents

```
Date: 10/2/95 at 2:31:12
From: Anonymous
Subject: Descartes' Method of Tangents

Given a point P(3,9) on a parabola y=x^2, how do I find the slope
of the tangent to the parabola at point P using Descartes method
of tangents.

Specifically, construct a point C(c,0) on the (+) x-axis and let
the line CP be the radius of a constructed circle. This circle
then has the equation (x-c)^2 + (y-0)^2 = (3-c)^2 + (9-0)^2.

So the problem is to find c so that (3,9) is the only solution to
both equations and then use this c to find the slope of both
tangent lines.

How does one do this?
```

```
Date: 10/2/95 at 13:11:41
From: Doctor Andrew
Subject: Re: Descartes' Method of Tangents

I would try solving for y in the equation of the circle.  Then
I would set up an equation to find the intersection of the
parabola and circle.  Then you could find c so that there is only
one point on that intersection, not two.  See if this helps.

-Doctor Andrew,  The Geometry Forum
```
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

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/