Tangents to a Curve That Pass through a Fixed PointDate: 08/08/2008 at 20:39:31 From: Sam Subject: tangent lines to y=x/(x+1) that pass through the point (1,2) How many tangent lines to the curve y = x/(x+1) pass through the point (1,2)? At which points do these tangent lines touch the curve? I'm not sure how to relate the derivative and the tangent lines in this problem. I already know from looking at a graph of this particular curve that there are two tangent lines that pass through the point (1,2). I also know that the derivative is y' = 1/[(x+1)(x+1)]. I know that the outputs of the derivative are the slopes of tangent lines at specified x values but I'm not sure how I use that information to find the tangent lines that pass through (1,2). I've set up a linear equation using the point-slope formula but I don't know how to include the derivative without confusing my variables. Please help me! Date: 08/10/2008 at 15:06:19 From: Doctor Ali Subject: Re: tangent lines to y=x/(x+1) that pass through the point (1,2) Hi Sam! Thanks for writing to Dr. Math. I'll try to explain two methods in brief. Assume that P(a, b) is not in y = f(x) and we want to find the lines tangent to the curve that pass through P. Method 1 -------- First we write the general equation of all lines, with any arbitrary slope, that pass through this special point P. That is, y - b = m(x - a) y = mx + b - ma Then, we want this line to be tangent to the curve. You can put y from above in the function to have, mx + b - ma = f(x) We know that the above equation has to have a double root. So not only should the above equality hold, but also the derivative of the sides should be equal. That is, m = f'(x) Do you see why? From here, you can find m's that satisfy the above conditions. Using those m's, you can write the equations of the lines. Method 2 -------- We know that these tangent lines touch the graph of the function at a certain point. Call that special point Q and assume Q(c, f(c)). Find the slope of the line that passes through P and Q. That is, f(c) - b m = ---------- c - a Now, we know that since the desired lines are tangent to the curve, the above slope has to be f'(c). Why c? This is because the x-coordinate of the point of tangency is c. So we can write, f(c) - b ---------- = f'(c) c - a In the above equality, we have everything except c. Solve the equation for c and write the equations of the lines using that. Now, can you apply either of the above methods to your own problem? Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ |
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