Using Descartes' Method of TangentsDate: 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 |
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