Finding the Equation of a Tangent LineDate: 11/24/98 at 13:57:19 From: Jennifer Ng Subject: Finding the equation of a tangent line (Calculus) I have looked through all the archives on how to find the equation of the tangent lines, but none could help me! In my case, instead of having the point where the tangent touches the main graph, I only have where the tangent intersects the y axis: "Line L is tangent to the graph of y=x-((x^2)/500) at the point Q. Find the equation of the tangent line by first finding the x-coordinate of point Q. Line L intersects the y-axis at point (0,20)" I have figured out that the derivative needs to be used in this case. When I finally find the x-coordinate, I will be able to put that into the derivative of the graph: 1-(x/250) and use the point-slope formula to find the equation. The problem is, I don't know the x-coordinate! Can you possibly help? Thanks! Jennifer Ng Date: 11/25/98 at 17:16:07 From: Doctor Nick Subject: Re: Finding the equation of a tangent line (Calculus) Hi Jennifer - One way to handle this is the following. Since we don't know what point on the curve the tangent line is tangent at, just call it (a,f(a)). Then what we want to do is determine what a has to be so that the tangent line passes through (0,20). The tangent line through (a,f(a)) has equation y-f(a) = f'(a) (x-a), i.e. y = (1-a/250)*(x-a)+(a-(a^2)/500)) which we can simplify to y = x*(1-a/250)+(a^2)/500. Thus, the y-intercept of this line is (a^2)/500. We want this to equal 20, so we need to solve 20 = (a^2)/500 which gives us a = (+ or -) sqrt(10000) = 100 or -100. So, in fact, there are _two_ lines that satisfy the requirements: one that's tangent at the point (100,80) and another at (-100,-120). This technique works for a lot of problems: if we don't know something (as here, the point of tangency), just give it a name (here (a,f(a))) and see where it leads us. You can get a long way with this method. Have fun, - Doctor Nick, The Math Forum http://mathforum.org/dr.math/ |
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