Box 4.5: Alternative Solution to Tangent Problem

Often there is more than one way to attack a problem. Such is the case with the problem of finding the lines tangent to

   x² - 2x + 1

that pass through the point, (5,7). Here is the solution that was, until recently, posted on Karl's Calculus Tutor, until a student emailed me with the more direct approach you see posted now.

Step 1: Find the derivative of the curve. In this case we have

   y'  =  2x - 2

We do this because we know that at the point of tangency, the derivative of the curve must be equal to the slope of the tangent line. So we need to know this derivative.

Step 2: Identify the information you get from the point. You are looking for the m and the b of the equation of a line, that is of

   y  =  mx + b

that passes through the point, (5, 7). That means that if you put in 5 for x and 7 for y, the above equation must work. That gives you

   7  =  5m + b

or equivalently

   b  =  7 - 5m

So wherever we see b later on, we can substitute 7 - 5m.

Step 3: Identify what happens at the points of tangency. You know that if the point, (x, y), is one of the points of tangency, then (x, y) must both lie on the curve, and at that point we know that m = y' = 2x - 2.

To find (x, y), we need to solve for both the x and the y in question. Since (x, y) lies on the curve, we know that

   y  =  x² - 2x + 1

holds at (x, y). So wherever we have a y, we can substitute x² - 2x + 1. And wherever we see an m, we can substitute 2x - 2.

Step 4: Make the substitutions into the equation for the line. Remember the equation for the line is y = mx + b. When you make the substitutions suggested by step 3, you get

   x² - 2x + 1  =  (2x - 2)x + b

But recall that from step 2, we have b = 7 - 5m. If you substitute for m, that becomes

   b  =  7 - 5(2x - 2)  =  17 - 10x

Putting it all together you have

   x² - 2x + 1  =  (2x - 2)x + 17 - 10x

Step 5: Gather like terms and solve for `x`. Using simple algebra the above equation becomes the quadratic x² - 10x + 16 = 0 You can either use the quadratic formula, or you can factor this one in your head to get (x - 2)(x - 8) = 0 So the `x` coordinate of the the point of tangency is either `x = 2` or `x = 8`. Step 6: Substitute back to get `m` and `b`. That is, we have already seen that `m = 2x - 2` at any point, `(x, y)`, of tangency. Since we now know what both the `x`'s are for the points of tangency, we can put those values into the equation for `m` and get that either `m = 2` or `m = 14`. We also know from step 2 that `b = 7 - 5m`. We have two values of `m` that solve the problem, so we plug those in to get that either `b = -3` or `b = -63`. So the equations of the two lines that are tangent to `y = x² - 2x + 1` and pass through the point, `(5, 7)`, are y = 2x - 3 and y = 14x - 63 and you are done. The curve and the two lines are illustrated in the graph on the right here. Notice where the two lines intersect. Both points of tangency are visible on the graph, although one is very nearly at the top of the graph.

Step 5: Gather like terms and solve for x. Using simple algebra the above equation becomes the quadratic

   x² - 10x + 16  =  0

You can either use the quadratic formula, or you can factor this one in your head to get

   (x - 2)(x - 8)  =  0

So the x coordinate of the the point of tangency is either x = 2 or x = 8.

Step 6: Substitute back to get m and b. That is, we have already seen that m = 2x - 2 at any point, (x, y), of tangency. Since we now know what both the x's are for the points of tangency, we can put those values into the equation for m and get that either m = 2 or m = 14. We also know from step 2 that b = 7 - 5m. We have two values of m that solve the problem, so we plug those in to get that either b = -3 or b = -63. So the equations of the two lines that are tangent to y = x² - 2x + 1 and pass through the point, (5, 7), are

   y  =  2x - 3

and

   y  =  14x - 63

and you are done. The curve and the two lines are illustrated in the graph on the right here. Notice where the two lines intersect. Both points of tangency are visible on the graph, although one is very nearly at the top of the graph.

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