MATH 680 : History of Mathematics

1 MATH 680 : History of Mathematics Term: Spring 013 Instructor: S. Furino Assignment 5: Analytic Geometry Weight: 8% Due: 3:55 (11:55PM) Waterloo time on 18 June 013 When typesetting your solutions use the document MATH680 S13 Assignment 5 Template.tex but replace the word Template with your family name. Family Name: Tapp First Name: Carrie I.D. Number: 0498109 If you used any references beyond the course text and lectures (such as other texts, discussions with colleagues or online resources), indicate this information in the space below. If you did not use any other references, state this in the space provided. University of San Francisco. (n.d.). Tangents to Curves. Retrieved on June nd, 013 from www.cs.usfca.edu/ cruse/math109s06/tangents.ppt

Mathematics 1. (30 marks) This question is meant to illustrate the difference between Greek geometric methods and the methods made available by Descartes descriptions of geometric curves as sets of points that satisfy an algebraic equation. (a) Given a point T on a parabola, give a sequence of steps for unmarked straightedge and compass that will construct a tangent to the parabola at T. Feel free to look this up using any reference you like. Step 1: Extend a line perpendicularly from the directrix to the point T. Call the point on the directrix, A. Step : Label M as the midpoint on a line from A to the focus, F. Step 3: Construct the line MT, this is the tangent line. (b) Prove that the line you constructed in (a) is indeed the tangent. Proof. T F = T A by the definition of a parabola, causing AF T to be isosceles. As M is the midpoint of AF, T M is a bisector and thus we end up with two identical triangles, T AM and T MF. As AF is a straight line split into two equal angles, T MA = T MF = 90. Due to the right angle, T M is a perpendicular bisector of AF. Extend the perpendicular bisector, T M, past T to a point,s. Connect a line perpendicuarly from the directrix, at a point called B, up to S. SAB will form a right triangle with hypotenuse SA and the right angle located at B. Thus SA > SB. SA = SF because both lines go to the same point on AF s perpendicular bisector axis. By substitution, SF > SB. As these two lines are not equal, S cannot be on the parabola with focus F and directrix on which A and B lie. Thus as there are no other points that lie on the parabola along the perpencidular bisector MT, MT is a tangent to the parabola. (c) Given the point T (8, 8), use the techniques of Descartes to find the equation of the tangent to y = ax at T. If y = ax has the point (8, 8) on it then a = 8 and y = 8x. The equation of the circle having its center on the x-axis and going through (8, 8) is Substituting y = 8x in (x x 1 ) + y = (8 x 1 ) + 8 (x x 1 ) + 8x = (8 x 1 ) + 8 x xx 1 + x 1 + 8x = 64 16x 1 + x 1 + 64 x xx 1 + 8x + 16x 1 + 18 = 0 x + (8 x 1 )x + 16(x 1 8) = 0 Comparing coefficients to (x r) = x rx + r = 0, r = 8 x 1 and r = 16(x 1 8). Making both equations equivalent to 4r by squaring the first and multiplying the second by 4, 4r = (8 x 1 ) = 64(x 1 8) 64 3x 1 + 4x 1 = 64x 1 51

3 4x 1 96x 1 + 576 = 0 4(x 1 4x 1 + 144) = 0 4(x 1 1)(x 1 1) = 0 x 1 = 1 The normal to the parabola at (8, 8) passes through (1, 0) providing an equation of y = x + 4. The required tangent will be perpendicular to this line, i.e. y = 1 x + 4. (d) Use calculus to find the equation of the tangent to y = ax at T (8, 8). Again, a = 8 yielding the equation y = 8x. Differentiating y = 8x provides yy = 8 y = 8 y As it is evaluated at T (8, 8) the slope at that point of the line is y = 8 16 = 1 Using point-slope form, y y 1 = m(x x 1 ) y 8 = 1 (x 8) y = 1 x 4 + 8 y = 1 x + 4

4. (10 marks) Burton 8.. Proof. Consider a point P between Q and R such that NP, of length b, is perpendicular to NL. As LM is a tangent, it will be perpendicular to NL and thus NP LM. As NR and NQ are radii, they are both of length a. Using the Pythagorean Theorem, ( a P R = NR NP = b ) = a 4 b Similarly, As MQ = MP P Q, As MR = MP + P R, P R = P Q = a a 4 b 4 b MQ = a a 4 b MR = a a + 4 b To show that MQ and MR are two positive solutions to x = ax b, it needs to be shown that both sides are equal when each length is set as x. With MQ, ( ) ( ) a a 4 a a b = a 4 b b a a 4 a 4 b + a 4 b = a a a 4 b b a a 4 a 4 b b = a a a 4 b b a a a 4 b b = a a a 4 b b As both sides are equal, MQ is a solution. With MR, ( ) ( ) a a + 4 a a b = a + 4 b b a a 4 + a 4 b + a 4 b = a a + a 4 b b a a 4 + a 4 b b = a a + a 4 b b a a + a 4 b b = a a + a 4 b b Again, as both sides are equal, MR is also a solution. As desired, MQ and MR are the positive solutions to x = ax b.

5 3. (10 marks) Burton 8..3 As line is the vertical axis, the distance of the point from line is only the horizontal x distance of the point. Hence, p = x. As line 5 is the horizontal axis, the distance of the point from line 5 is only the vertical y distance of the point. Hence p5 = y. Line 1 is to the left of Line so it s distance to the point will include the x distance to line plus the distance between the lines, which is called a. Hence p1 = a + x. Line 3 is to the right of line and as the point is inbetween Line and Line 3, the distance from line 3 to the point is the distance between line and line 3, minus the distance from the point to line. Hence, p3 = a x. Line 4 is an extra distance of a from the point p than line 3 was, hence p4 = a x + a = a x. By substiution this provides the equation of (a + x)(a x)(a x) = axy for the locus of points satisfying p1 p p4 = ap p5.

6 4. (10 marks) Burton 8.4.13 a) The first requirement is to show that triangles BPC and OAB are similar, whence (BP)(AB)=(OA)(PC). As P is the intersection of a parallel and perpendicular line, the measurement of BP C is 90. A is also at the intersection of the parallel tangent and the perpendicular vertical axis, providing OAB =90. Due to the rules of a transversal through two parallel lines, ABO = BCP and AOB = P BC by alternate interior angles. As there are three equal pairs of angles, triangles BP C and OAB are similar. b) The second requirement is to show that If O is the origin of a rectangular coordinate system and P and C have coordinates (x, y) and (u, v), respectively, then (a y)x = a(x u) and so u = (xy)/a, while v = y. From part a, it is known that (BP )(AB) = (OA)(P C). When y = v substitution based on the new coordinate labelling provides the desired (a y)x = a(x u). Distribution yields ax yx = ax au. Cancelling like terms gives yx = au and solving for u shows the desired u = xy a. c) The final step is to substitute u = xy a centre (0, a/). and v = y into an equation of a circle with radius, a/ and ( u + v a ) ( a = ) ( xy ) ( + y a ) ( a = a ) x y a + y ay + a 4 = a 4 x y a + y ay = 0 x y a + y a = 0 x y a + y = a x y + a y = a 3 y(x + a ) = a 3 As desired, the locus of point P as the secant varies is the desired curve with equation y(x +a ) = a 3.

7 5. (0 marks) Read pages 373 374 which describe how Descartes solved quartic equations. Using this technique, find the roots of x 4 3x + 6x = 0 (Hint: k = 4.) Following Descartes notation, p = 3, q = 6, and r =. Rewriting in cubic form provides k 6 6k 4 + 17k 36 = 0 As k = 4 makes the above statement true, the following pair of quadratics now need to be solved with k = 4 and k = ±. z + kz + 1 ( 3 + k 6 ) = 0 k z kz + 1 ( 3 + k + 6 ) = 0 k Choosing k = so as to avoid duplicate solutions, simplifies these to z + z + 1 ( 3 + 4 6 ) = z + z 1 = 0 z z + 1 ( 3 + 4 + 6 ) = z z + = 0 Via the quadratic formula, the real roots stemming from the first equation are 1 + and 1 and the imaginary roots coming from the second equation are 1 + i and 1 i.

8 6. (0 marks) The 1998 Euclid contest is posted online at http://www.cemc.uwaterloo.ca/contests/ past_contests/1998/1998euclidcontest.pdf. Without checking the online solutions, answer Question 8 of the 1998 Euclid contest. Proof. As the parabola is only translated, not transformed by a stretch or compression, the new equation in standard form will be y = x + bx f. Using the factors of the parabola, the equation could also be written as y = (x + d)(x e) = x + (d e)x de. As both equations are of the same parabola, by substitution x + bx f = x + (d e)x de and it becomes apparent that f = de. Part (b) follows on the next page.

9 Proof. In order to prove that WA is parallel to KD, one could show that they have the same slopes. Label the points on a coordinate plane as follows: K(0, 0), M(a, b), W (a, b), A(c, d), D(e, 0), N(c+ e, b). This labelling ensures that M and N are midpoints of their respective lines. The slope of KD is 0 0 e 0 = 0 so it needs to be shown that the slope of W A = 0, i.e. d b c a = d b c a = 0 and as the only way to make a fraction 0 is to have a 0 in the denominator, it ultimately needs to be shown that d b = 0. As MN = 1 (AW + DK) = 1 AW + 1 DK, using distance formula gives: (c + e a) + (d b) = 1 (c a) + (d b) + 1 (e 0) + (0 0) (c + e a) + (d b) = 1 (c a) + (d b) + e Squaring both sides, (c + e a) + (d b) = 1 ( (c a) + (d b) ) + e (c a) 4 + (d b) + e Distributing, c +ce ac+e +a ea+(d b) = 1 ( 4c 8ac + 4a ) + 1 4 4 ((d b)) +e ((c a)) + ((d b)) +e c + ce ac + e + a ea + (d b) = c ac + a + (d b) + e ((c a)) + ((d b)) + e Canceling, ce ae = e ((c a)) + ((d b)) (c a) = ((c a)) + ((d b)) Squaring both sides, 4(c a) = 4(c a) + 4(d b) (c a) = (c a) + (d b) Thus, d b must equal 0, providing a slope of 0 for W A and therefore, W A KD.