Parabolas and lines
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1 Parabolas and lines Study the diagram at the right. I have drawn the graph y = x. The vertical line x = 1 is drawn and a number of secants to the parabola are drawn, all centred at x=1. By this I mean that the x-intervals defining the secants are all centred at x=1. These are the intervals [0, ], [ 1, 3] and [, 4] What do you notice about these secants? They are all parallel. They all have the same slope! Does that surprise you? Maybe you think it s quite normal, even expected? But it s rare. For example at the right I draw the same set of secants for the cubic polynomial y = x (6 x) These secants are far from parallel. In fact the parabola is the only curve that has that property for every x. That is, if you draw a curve that has this property for every x, it will be a parabola. Let s do the algebra. Problem 1. Return to the parabola y = x. The intervals centred at x=1 all have the form [1 d, 1+d] for some d. Show algebraically that the secants on these intervals all have the same slope, that is, that the slope is independent of d. Solution. The slope of the secant is rise f(1 + d) f(1 d) = = (1 + d) (1 d) run (1 + d) (1 d) d + d = (1 + d + d ) (1 d + d ) d = 4d d = and that s independent of d. So all secants centred at x = 1 have slope. peter.taylor@queensu.ca 1
2 Problem. Having shown that secants centred at x=3 are all parallel, is this also the case for any other x? Do the secants centred at x=a all have the same slope and what is this slope Solution. The slope of the secant on [a d, a+d] rise run = (a + d) (a d) (a + d) (a d) = (a + ad + d ) (a ad + d ) d = 4ad d = a And that s again independent of d. So all secants centred at x=a have slope a. For example, the secants centred at x= at the right all have slope 4. peter.taylor@queensu.ca
3 Parabola and line translation At the right I have drawn the graphs of the parabola And the line y = x(6 x) y = x Note that the line intersects the parabola twice, at x = 0 and x = 4. What I want to do is translate the line upwards, parallel to itself, keeping track of the number of intersections with the parabola. If we do this graphically, we see that at the beginning there ae always two intersections, but at a certain point the line is tangent to the parabola with only one intersection, and if we move the line above that, there are no intersections. The point of tangency seems to be at x =. Problem 3. Track this translation process algebraically. That is, find an algebraic equation to describe this family of lines and use it to check the details of the intersection story. Solution. I begin by putting subscripts on the variable y to distinguish the two different functional dependences, the parabola and the starting line. y P = x(6 x) y L = x As a warm-up for what lies ahead, let s see what we need to do with these two equations to find where the two graphs intersect. Well, they intersect at points where the two y-values are equal: y L = y P x = x(6 x) Cancel the x s: = 6 x x = 4 That s one of the intersections but there s another at the origin. We lost that one when we canceled the x s. It might have been safer not to do that: x = x(6 x) = 6x x x 4x = 0 x(x 4) = 0 x 0, x = 4 and this way we get the two solutions. peter.taylor@queensu.ca 3
4 Now we look at the family of lines. As we move the line up, we increase the y-intercept b so that b is a ready-made parameter to distinguish the different lines. The family is: y L = x + b Notice the terminology. We call x and y variables and we call b a parameter. In a sense they are all variables because they all vary. But we think of them in different ways x and y refer to the axes and they tell us how the point moves along any one of the lines, whereas b is like an index that keeps track of the different lines. The term parameter is a signal that we have an indexed family of objects (lines). As above, we are going to intersect y L and y P but this time there are many lines and we expect different intersections for different members of the family (different values of b). Just to review, when b is just above 0, we expect intersections; as we increase b these two intersections get closer together and at some point (looks like b = 4) they merge into one, and for b above 4, we have no solutions. Let s see if the algebra gives us that. As before we equate the y-expressions Collect powers of x and put everything on the left: y L = y P x + b = 6x x x 4x + b = 0 We want the solutions of this equation the values of x for which it holds. The quadratic polynomial does not factor so we use the quadratic formula: 4 ± 16 4b x = The precise question we are interested in is this: for each particular b, how many solutions do we have? The key lies in the discriminant, the expression under the root sign. If this is positive, the sign will give us two solutions, if it s zero, we ll just get one, and if it s negative, we won t get any. Thus: Solving for b: if 16 4b > 0 if 16 4b = 0 if 16 4b < 0 b < 4 b = 4 b > 4 two solutions one solution no solutions. we have two solutions we have one solution we have no solutions. and this fits exactly with our graphical solution. peter.taylor@queensu.ca 4
5 Parabola and line rotation At the right I have drawn the graph of the parabola y P = x(6 x) along with a family of lines passing through the point (5, 5). Again I am interested in the pattern of intersections of the lines with the parabola. Clearly they all intersect the parabola at (5, 5). Some of them clearly have another intersection, for example one of them has an intersection at the origin. I guess it s clear that the vertical line x = 5 only has the one intersection (is it?). What s the general story? How many lines have no other intersection (other than (5, 5))? Problem 4. Analyze this configuration algebraically. Find an algebraic equation to describe the family of lines and use it to answer the questions posed above. First we need an equation for the family of lines passing through (5, 5). Since they all have different slopes, the slope m should provide a good parameter. For those students who remember the point-slope form for the equation of a line, the answer is simple: y 5 = m(x 5) But most of my students do not naturally use that form. Instead they use y = mx + b and then find b by plugging in the point (5, 5): 5 = 5m + b 5 = 5m + b b = 5(1 m) By either of the two methods, the family is: y L = mx + 5(1 m) What s the intersection story? Which lines have no intersection other than (5, 5)? It s a good discussion. Someone mentions the tangent line. Others tell me that lines that are very nearly vertical won t intersect the parabola again. Now we intersect each member of this family with the parabola. Set the heights y equal: y L = y P mx + 5(1 m) = 6x x Collect powers of x and put everything on the left: x (6 m)x + 5(1 m) = 0 We want the solutions of this equation the values of x for which it holds. The quadratic polynomial does not factor so we use the quadratic formula: peter.taylor@queensu.ca 5
6 x = 6 m ± (6 m) 0(1 m) Recall that the question we are interested in is this: for each particular slope m, how many solutions do we have? As before, the key lies in the discriminant. If this is positive, the sign will give us two solutions, if it s zero, we ll just get one, and if it s negative, we won t get any. Thus we want the sign of (6 m) 0(1 m) = 36 1m + m 0 + 0m = m + m(0 1) = m + 8m + 16 = (m + 4) Goodness. The discriminant is a perfect square. It is always bigger than or equal to zero, and it is zero only when: (m + 4) = 0 m = 4 This is a negative slope and is clearly the slope of the tangent at (5, 5). The interesting conclusion is that all other lines in the family intersect the parabola twice. What happened to the vertical line? Well it was omitted from the analysis. It is not of the form y = mx + b. Indeed what is the equation of the vertical line? Interestingly this question of mine meets a blank stare. It is an unusual question for them. Finally someone produces it: x = 5 peter.taylor@queensu.ca 6
7 Problems 1. On the same set of axes, draw the graph of the parabola y = x and the line y = 4x. Observe that the line intersects the parabola exactly twice, once at x=0 and once at x=4. What you are to do is imagine the family of all lines of slope 4 and ask yourself the question how many intersections does each line have with the parabola? (a) Study your picture and make a conjecture which answers the question. (b) Describe the family of lines algebraically, and verify the correctness of your conjecture. Illustrate your solution with a picture. (c) Use the result of (b) to find the value of x at which the tangent to the curve has slope 4.. At the right is the graph of the curve and the line y = 1/x y = x/4. Observe that the line fails to intersect the curve. What you are to do is imagine the family of all lines of slope 1/4 and ask yourself the question how many intersections does each line have with the curve? (a) Study the picture and make a conjecture which answers the question. (b) Describe the family of lines algebraically, and verify the correctness of your conjecture. Illustrate your solution with a picture. (c) Use the result of (c) to find two values of x at which the tangent to the curve has slope 1/4. peter.taylor@queensu.ca 7
8 3. At the right is the graph of the parabola y = (x 1)(4 x) together with a line which is tangent to the curve and passes through the origin. (a) Find the slope of the line. [Hint: You can see the answer, m = 1, from the diagram. But here you are asked to use the algebra. Work with the family of all lines passing through the origin.] (b) Is there another line through the origin which is tangent to the curve? If so, what is its slope? 4.(a) Find in terms of b the number of intersections of the curve with the line x y 1 x y = x+b. (b) Use the above to find all values of x at which the curve has slope 1. (c) Perhaps you've done this problem "blind" so far. If so, now is the time to try to illustrate your results on a sketch of the graph. Now you may find the prospect of drawing the graph y = x/(1+x) a bit daunting. But # above contains a picture of y=1/x and you might note that this graph is closely related all you need are translations and sign changes. Indeed: Can you take it from here? x x x 1 x 1 x 5. Find in terms of b the number of intersections of the circle x y. with the line y = b x. Draw a picture to illustrate your solution and show that it is geometrically expected. 6. Sketch a number of members of the family of parabolas y k ( x 1), indexed by k. In terms of k, how many intersections are there with the parabola y x? In particular, for what k are the curves tangent? Use your diagram to illustrate your solution. peter.taylor@queensu.ca 8
9 7. How do different members of the family x y 1 y x k of parabolas intersect the circle? Before you do the algebraic analysis, use your intuition to try to guess the form of the answer. Illustrate your solution with a picture. 8. In terms of the parameter a, when does the pair of equations x y = 0 have 0, 1,, 3, 4, or 5 solutions? (x a) +y = 1 [It s important to get the right picture. The first equation is simpler than it looks write it as x = y and take square roots. And the second equation is a moving circle.] 9. Find the radius of the largest circle that lies above the x-axis and underneath the parabola. [Answer: r = 3/] y 4 x The last 4 problems are all challenging, and any one of them can make a great project. #8 is an interesting problem because it appears in A Source Book for College Mathematics Teaching (Mathematical Association of America 1990, Alan Schoenfield, Ed.). It is said to be an example of a problem with which most students, armed with the traditional calculus skills, will not really manage to make much headway. But that's not us 10. Find the value of k for which the parabolas y x k and are tangent. [Answer: k = 3/(4 4/3 ) ] x y peter.taylor@queensu.ca 9
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