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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

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