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MATH 680 : History of Mathematics

Term: Spring 2013

Instructor: S. Furino

Assignment 5: Analytic Geometry

Weight: 8%

Due: 23:55 (11:55PM) Waterloo time on 18 June 2013

When typesetting your solutions use the document MATH680 S13 Assignment 5 Template.tex but replacethe word “Template” with your family name.

Family Name: Tapp

First Name: Carrie

I.D. Number: 20498109

If you used any references beyond the course text and lectures (such as other texts, discussions withcolleagues or online resources), indicate this information in the space below. If you did not use any otherreferences, state this in the space provided.

University of San Francisco. (n.d.). Tangents to Curves. Retrieved on June 2nd, 2013 fromwww.cs.usfca.edu/ cruse/math109s06/tangents.ppt

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Mathematics

1. (30 marks) This question is meant to illustrate the difference between Greek geometric methods andthe methods made available by Descartes’ descriptions of geometric curves as sets of points thatsatisfy an algebraic equation.

(a) Given a point T on a parabola, give a sequence of steps for unmarked straightedge and compassthat will construct a tangent to the parabola at T . Feel free to look this up using any referenceyou like.

Step 1: Extend a line perpendicularly from the directrix to the point T . Call the point on thedirectrix, A.Step 2: 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. TF = TA by the definition of a parabola, causing 4AFT to be isosceles.As M is the midpoint of AF , TM is a bisector and thus we end up with two identical triangles,4TAM and 4TMF . As AF is a straight line split into two equal angles, ∠TMA = ∠TMF =90◦. Due to the right angle, TM is a perpendicular bisector of AF .Extend the perpendicular bisector, TM , past T to a point,S.Connect a line perpendicuarly from the directrix, at a point called B, up to S.4SAB 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 perpendicularbisector axis.By substitution, SF > SB.As these two lines are not equal, S cannot be on the parabola with focus F and directrix onwhich 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 toy2 = ax at T . If y2 = ax has the point (8, 8) on it then a = 8 and y2 = 8x. The equation of thecircle having its center on the x-axis and going through (8, 8) is

(x− x1)2 + y2 = (8− x1)

2 + 82

Substituting y2 = 8x in(x− x1)

2 + 8x = (8− x1)2 + 82

x2 − 2xx1 + x21 + 8x = 64− 16x1 + x21 + 64

x2 − 2xx1 + 8x + 16x1 + 128 = 0

x2 + (8− 2x1)x + 16(x1 − 8) = 0

Comparing coefficients to (x− r)2 = x2 − 2rx + r2 = 0, −2r = 8− 2x1 and r2 = 16(x1 − 8).Making both equations equivalent to 4r2 by squaring the first and multiplying the second by 4,

4r2 = (8− 2x1)2 = 64(x1 − 8)

64− 32x1 + 4x21 = 64x1 − 512

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4x21 − 96x1 + 576 = 0

4(x21 − 24x1 + 144) = 0

4(x1 − 12)(x1 − 12) = 0

x1 = 12

The normal to the parabola at (8, 8) passes through (12, 0) providing an equation ofy = −2x + 24. The required tangent will be perpendicular to this line, i.e. y = 1

2x + 4.

(d) Use calculus to find the equation of the tangent to y2 = ax at T (8, 8).Again, a = 8 yielding the equation y2 = 8x. Differentiating y2 = 8x provides

2yy′ = 8

y′ =8

2y

As it is evaluated at T (8, 8) the slope at that point of the line is

y′ =8

16=

1

2

Using point-slope form,y − y1 = m(x− x1)

y − 8 =1

2(x− 8)

y =1

2x− 4 + 8

y =1

2x + 4

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2. (10 marks) Burton 8.2.2

Proof. Consider a point P between Q and R such that NP , of length b, is perpendicular to NL. AsLM 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

2 .Using the Pythagorean Theorem,

PR2 = NR2 −NP 2 =(a

2

)2− b2 =

a2

4− b2

PR =

√a2

4− b2

Similarly,

PQ =

√a2

4− b2

As MQ = MP − PQ,

MQ =a

2−√

a2

4− b2

As MR = MP + PR,

MR =a

2+

√a2

4− b2

To show that MQ and MR are two positive solutions to x2 = ax − b2, it needs to be shown thatboth sides are equal when each length is set as x. With MQ,(

a

2−√

a2

4− b2

)2

= a

(a

2−√

a2

4− b2

)− b2

a2

4− 2

a

2

√a2

4− b2 +

a2

4− b2 =

a2

2− a

√a2

4− b2 − b2

2a2

4− a

√a2

4− b2 − b2 =

a2

2− a

√a2

4− b2 − b2

a2

2− a

√a2

4− b2 − b2 =

a2

2− a

√a2

4− b2 − b2

As both sides are equal, MQ is a solution.

With MR, (a

2+

√a2

4− b2

)2

= a

(a

2+

√a2

4− b2

)− b2

a2

4+ 2

a

2

√a2

4− b2 +

a2

4− b2 =

a2

2+ a

√a2

4− b2 − b2

2a2

4+ a

√a2

4− b2 − b2 =

a2

2+ a

√a2

4− b2 − b2

a2

2+ a

√a2

4− b2 − b2 =

a2

2+ a

√a2

4− b2 − b2

Again, as both sides are equal, MR is also a solution.

As desired, MQ and MR are the positive solutions to x2 = ax− b2.

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3. (10 marks) Burton 8.2.3

As line 2 is the vertical axis, the distance of the point from line 2 is only the horizontal x distance ofthe point. Hence, p2 = x.As line 5 is the horizontal axis, the distance of the point from line 5 is only the vertical y distance ofthe point. Hence p5 = y.Line 1 is to the left of Line 2 so it’s distance to the point will include the x distance to line 2 plusthe distance between the lines, which is called a. Hence p1 = a + x.Line 3 is to the right of line 2 and as the point is inbetween Line 2 and Line 3, the distance from line3 to the point is the distance between line 2 and line 3, minus the distance from the point to line 2.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 = 2a− x.By substiution this provides the equation of (a + x)(a − x)(2a − x) = axy for the locus of pointssatisfying p1p2p4 = ap2p5.

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4. (10 marks) Burton 8.4.13a) 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 ∠BPC 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 = ∠PBCby alternate interior angles.As there are three equal pairs of angles, triangles BPC and OAB are similar.

b) The second requirement is to show that “If O is the origin of a rectangular coordinate system andP 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)(PC). When y = v substitution based on the newcoordinate 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 ushows the desired u = xy

a .

c) The final step is to substitute u = xya and v = y into an equation of a circle with radius, a/2 and

centre (0, a/2).

u2 +(v − a

2

)2=(a

2

)2(xya

)2+(y − a

2

)2=(a

2

)2x2y2

a2+ y2 − 2

ay

2+

a2

4=

a2

4

x2y2

a2+ y2 − ay = 0

x2y

a2+ y − a = 0

x2y

a2+ y = a

x2y + a2y = a3

y(x2 + a2) = a3 �

As desired, the locus of point P as the secant varies is the desired curve with equation y(x2+a2) = a3.

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5. (20 marks) Read pages 373 – 374 which describe how Descartes solved quartic equations. Using thistechnique, find the roots of

x4 − 3x2 + 6x− 2 = 0

(Hint: k2 = 4.)

Following Descartes’ notation, p = −3, q = 6, and r = −2. Rewriting in cubic form provides

k6 − 6k4 + 17k2 − 36 = 0

As k2 = 4 makes the above statement true, the following pair of quadratics now need to be solvedwith k2 = 4 and k = ±2.

z2 + kz +1

2

(−3 + k2 − 6

k

)= 0

z2 − kz +1

2

(−3 + k2 +

6

k

)= 0

Choosing k = 2 so as to avoid duplicate solutions, simplifies these to

z2 + 2z +1

2

(−3 + 4− 6

2

)= z2 + 2z − 1 = 0

z2 − 2z +1

2

(−3 + 4 +

6

2

)= z2 − 2z + 2 = 0

Via the quadratic formula, the real roots stemming from the first equation are −1 +√

2 and −1−√

2and the imaginary roots coming from the second equation are 1 + i and 1− i.

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6. (20 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 Ques-tion 8 of the 1998 Euclid contest.

Proof. As the parabola is only translated, not transformed by a stretch or compression, the newequation in standard form will be y = x2 + bx− f .Using the factors of the parabola, the equation could also be written as y = (x + d)(x − e) =x2 + (d− e)x− de.As both equations are of the same parabola, by substitution x2 + bx− f = x2 + (d− e)x− de and itbecomes apparent that f = de.

Part (b) follows on the next page.

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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 (2a, 2b), A(2c, 2d), D(2e, 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

2e−0 = 0 so it needs to be shown that the slope of WA = 0, i.e. 2d−2b2c−2a = d−b

c−a = 0and as the only way to make a fraction 0 is to have a 0 in the denominator, it ultimately needs to beshown that d− b = 0.

As MN = 12(AW + DK) = 1

2AW + 12DK, using distance formula gives:√

(c + e− a)2 + (d− b)2 =1

2

√(2c− 2a)2 + (2d− 2b)2 +

1

2

√(2e− 0)2 + (0− 0)2√

(c + e− a)2 + (d− b)2 =1

2

√(2c− 2a)2 + (2d− 2b)2 + e

Squaring both sides,

(c + e− a)2 + (d− b)2 =1

4

((2c− 2a)2 + (2d− 2b)2

)+ e√

(2c− 2a)2 + (2d− 2b)2 + e2

Distributing,

c2+2ce−2ac+e2+a2−2ea+(d−b)2 =1

4

(4c2 − 8ac + 4a2

)+

1

4(2(d− b))2+e

√(2(c− a))2 + (2(d− b))2+e2

c2 + 2ce− 2ac+ e2 + a2− 2ea+ (d− b)2 = c2− 2ac+ a2 + (d− b)2 + e√

(2(c− a))2 + (2(d− b))2 + e2

Canceling,2ce− 2ae = e

√(2(c− a))2 + (2(d− b))2

2(c− a) =√

(2(c− a))2 + (2(d− b))2

Squaring both sides,4(c− a)2 = 4(c− a)2 + 4(d− b)2

(c− a)2 = (c− a)2 + (d− b)2

Thus, d− b must equal 0, providing a slope of 0 for WA and therefore, WA ‖ KD.