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Tangents to a Parabola

Date: 10/21/96 at 23:39:23
From: Dave Spoelhof
Subject: Finding tangents to parabolas

I am trying to help my 10th grader do a math problem. I think this may 
be simple for someone who does a lot of algebra and calculus.  The 
question is to give the slope of two lines which are tangent to the 
parabola y = x^2 and pass through the point (2,1).  We figured that 
the line between (2,1) and (0,0) with a slope of 1/2 was one answer, 
but we are stumped trying to figure the slope of a second line that is 
tangent and passes through (2,1).  Can you help?

Dave Spoelhof

Date: 10/22/96 at 2:59:12
From: Doctor Pete
Subject: Re: Finding tangents to parabolas

If your child has not had a course in calculus, then this problem is
somewhat difficult to solve because you need the fact that the slope 
of the line tangent to the parabola y = x^2 at some point x is 2x 
(this is essentially the meaning of the derivative).  Given this fact, 
take some point (a,y) = (a,a^2) on the parabola.  The line through 
this point and tangent to the parabola has slope 2a, so its equation 
is y = 2ax + b where b is the y-intercept.  We find b by noting that 
the point (a,a^2) is on the line, thus a^2 = 2a(a) + b, or b = -a^2.  
Hence the equation becomes:

     y = 2ax - a^2.

It is important at this point to realize that this is not the equation 
of one particular line, but rather an entire *family* of lines, one 
for each possible value of a.  Thus, we want to find the values of a 
for which the above equation describes a line passing through the 
point (2,1).  To this end, we simply substitute these coordinates into 
the equation and solve:

     1 = 2a(2) - a^2
     0 = a^2 - 4a + 1
     a = {(4+Sqrt[16-4])/2, (4-Sqrt[16-4])/2}
       = {2+Sqrt[3], 2-Sqrt[3]}

Therefore, there are exactly two lines which satisfy the requirements 
of the problem. They are:

     y = (4+2*Sqrt[3])x - (7+4*Sqrt[3])

     y = (4-2*Sqrt[3])x - (1-4*Sqrt[3])

An observation is that the line you described as passing through (2,1) 
and (0,0), cannot be tangent to the parabola y = x^2. This is because 
the tangent to the parabola at (0,0) is the line y = 0 (i.e., the x-
axis); since (2,1) is not on this line and since every point on the 
parabola has a unique tangent line, the line described cannot be 
tangent.  Indeed, the line y = x/2 intersects the parabola y = x^2 at 
two points: (0,0) and (1/2, 1/4).

To show that the slope of the tangent line to y = x^2 at some point x 
is 2x, consider the following argument.  If you have two points 
P1 = (a,a^2) and P2 = (a+h,(a+h)^2) on the parabola, where h is small 
compared to a, then the line passing through P1 and P2 is a reasonable 
approximation to the tangent line through (a,a^2).  Furthermore, one 
would expect that as h tends to 0, the the line through P1 and P2 gets 
closer and closer to the tangent line at P1.  In fact, they coincide 
at the *limit* as h -> 0.  Thus, the slope of the tangent line is 
reasonably approximated by:

                (a+h)^2 - a^2   a^2 + 2ah + h^2 - a^2
     m(P1,P2) = ------------- = ---------------------
                 (a+h) - a                h

                2ah + h^2
              = --------- = 2a + h.

But as h -> 0, m(P1,P2) -> 2a (that is, as h tends to 0, the slope 
tends to 2a).  Strictly speaking, the *limit* as h goes to 0 *is* 2a.  
Therefore, the slope of the tangent to y = x^2 at some point (a,a^2) 
is 2a.

I would consider the above to be fairly advanced for a 10th grader 
since it introduces the concepts of slope as well as the equivalence 
between intersections of lines and curves to solutions of systems of 
equations, limits of functions, and the geometric interpretation of 
derivatives as tangents to curves.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Calculus

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