Tangents to a ParabolaDate: 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? Thanks, 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. 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 Check out our web site! http://mathforum.org/dr.math/ |
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