Descartes' Method for Tangents
Date: 09/02/98 at 13:38:38 From: Sara Subject: Finding the equation of the tangent line using Descartes' method I've been pretty puzzled by this one. It is really confusing mostly because I have never done a problem like this before because I've only taken Pre-calc. This is my Freshman year in college and this is a Calc I problem. When you do it could you please explain every step and why you did it please? I will write out the problem for you. Luckily you can figure out what the picture looks like if you read carefully. Rene Descartes (1596-1650) had the following solution to the construction of tangent lines: When given the equation of a curve, say the parabola y^2 = 2x, to construct the tangent line to the curve at the point (2,2), we will look at the family of all circles whose center (a,0) is on the x-axis and which passes through the point (2,2). Only one of these circles will intersect the parabola at only one point (i.e. be tangent to the parabola). When we find this circle, Descartes surmised, then the tangent to that circle will be the tangent to the parabola at (2,2). Using this method, find the equation of the tangent line.
Date: 09/02/98 at 16:43:55 From: Doctor Rob Subject: Re: Finding the equation of the tangent line using Descartes' method The equation of a circle whose center is at (a,0) is: (x-a)^2 + y^2 = r^2 If the circle passes through (2,2), then you know that: (2-a)^2 + 2^2 = r^2 r^2 = a^2 - 4*a + 8 Thus the family of all these circles will have equations: (x-a)^2 + y^2 = a^2 - 4*a + 8 The parabola has equation y^2 = 2*x. The points of intersection of the parabola and the circle will satisfy both equations, so solve them simultaneously. The fastest way is to substitute y^2 = 2*x in the equation for the circle: (x-a)^2 + 2*x = a^2 - 4*a + 8 x^2 + (2-2*a)*x + (4*a-8) = 0 Since x = 2 is a solution to this equation, no matter the value of a, x - 2 must be a factor of the lefthand side. Sure enough, this last equation is the same as: (x - 2)*(x - 2*a + 4) = 0 so x = 2*a - 4 must be the x-coordinate of the other point of intersection. For these two points to coincide, you must have 2 = 2*a - 4, so a = 3. Another way to find a is to realize that for the quadratic in x above to have two equal roots, 0 = B^2 - 4*A*C = (2-2*a)^2 - 4*(4*a-8) = 4*(a-3)^2, so a = 3. Now all you have to do is find the tangent to the circle: (x-3)^2 + y^2 = 5 at the point (2,2). I leave this to you. There are several ways to do this. One uses the derivative dy/dx. Another uses the fact that the tangent line is perpendicular to the radius, which connects (2,2) to (3,0). - Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.