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Descartes' Method for Tangents

Date: 09/02/98 at 13:38:38
From: Sara
Subject: Finding the equation of the tangent line using Descartes' 

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' 

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 

- Doctor Rob, The Math Forum
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Associated Topics:
High School Calculus
High School Equations, Graphs, Translations

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