Figure 1: ET is an oblique-angled diameter

1 Snopsis b section of Euler s paper ON SOME PROPERTIES OF CONIC SECTIONS THAT ARE SHARED WITH INFINITELY MANY OTHER CURVED LINES (E08) Thomas J Osler and Edward Greve Mathematics Department Rowan Universit Glassboro, NJ 0808 Osler@rowanedu Introduction Euler wrote of oblique-angled diameters and orthogonal diameters without explanation A E M N C T D B Figure 1: ET is an oblique-angled diameter Definition: Given a curve ACTDB shown in figure 1 The line ET, which intersects the curve at T, is called an oblique-angled diameter if it bisects all chords (such as AB and CD), that are parallel to the tangent line at T

In other words AM MB and CN ND since the chords AB and CD are parallel to the tangent line at the point T where the oblique-angled diameter ET intersects the curve ACTDB Part I: Curves with parallel diameters Section 1 We will examine the concept of oblique-angled diameters, which applies to the conic sections, and tr to find other curves with that propert or a similar propert Figure 1 Section Euler reminds us that an line parallel to the axis of a parabola is an obliqueangled diameter Section We now stud the following problem: To describe a curve AMFm over the axis AD, which has the diameter FEG parallel to the axis AD at the given distance DE from the axis, and which bisects all chords Mm which make (a fixed) angle β at T with the axis

With the notation shown in Figure 1, and with n sin β and m cos β, Euler reasons that z P( t) z Q( t ) Here Pt () and Qt () are arbitrar functions of t (Notice that the letters m and P have ambiguous meanings Euler does this frequentl) Section 4 Call z M and the roots of the equation z P( t) z Q( t) Using z m Pt ( ) zm + zm and simple reasoning from the geometr of Figure 1, Euler concludes that the equation of the curve AMFm is a f( n mx), where the function f is arbitrar Section 5 We set n mx X and a Y and form the general equation between X and Y, b writing 0 α βx γy δx εxy ζy ηx θx Y + + + + + + + + (1) In this general equation we find all possible relations between X and Y and thus we have Y to an function of X, so that a will be equal to an function of n mx as required b our analsis Section 6 When the curve is a parabola, all lines parallel to the axis are oblique-angled diameters, while in the famil of curves we found, onl one line (at distance a from the axis) is an oblique-angled diameter Section 7 Now Euler considers curves that are smmetrical about the axis He poses the problem: Among all curves AMm, with the line AD as an axis of smmetr, to determine those that at a given distance on both sides of the axis AD have two oblique-angled diameters Because of the smmetr, the variable cannot occur to an odd power in the equation of the curve But because both X n mx and Y afeature to the

4 first power, Euler seeks new variables without odd powers He selects ax m a x Z Y n n + He notes that the preceding problem (without smmetr) is solved b the general equation between Y and Z, that is to sa: 0 α + βy + γz + δy + εyz + ςz + ηy + θy Z + () Section 8 Because the variable Y contains to the first power, our second problem, with axis of smmetr, can onl be satisfied if terms involving odd powers of are eliminated Euler argues that the removal of odd powers of requires the use of infinitel man terms involving 5 Y, Y, Y, Thus if we restrict ourselves to algebraic equations, (finitel man terms), equation () must reduce to q 0 α + γz + ςz + χz + + ωz, that contains no Y It follows that Z is a constant max C n This is the parabola, and therefore all other algebraic curves are excluded Section 9 While the onl algebraic equations with an axis of smmetr and obliqueangled diameters are parabolas, there ma be man transcendental curves with this propert Section 10 While Euler will not pursue transcendental curves further, he indicates a method for finding them With Y a and Z ma x we seek a function of Y n and Z with no odd powers of, since the curve is smmetric about the x axis Euler asks

5 us, without explanation, to find a function TY ( ) satisfing the infinite differential equation 4 16 64 0 + + + + () 7 dt ayd T 4 5 a Y d T 6 7 a Y d T dy 1 dy 5 145dY 17dY He claims that this function T has no odd powers of With W ( T, Z) an arbitrar function, the equation WTZ (, ) 0 is a transcendental curve with axis of smmetr and oblique-angled diameter(s) We show our derivation of () in the notes Section 11 Euler gives a neat, simple geometric argument to prove the following theorem We redo this proof in the notes with more details than given b Euler Theorem If a curve has two parallel oblique-angled diameters, separated b the distance a, then that curve has infinitel man parallel oblique-angled diameters, all separated b the same distance a Section 1 The figure shows a curve with oblique-angled diameters Aa, Bb, Cc and corresponding angles α, β, γ From the previous section we know that there are infinitel man diameters all equall spaced b the distance a Referring to the figure we will use θ1 α, the angle made b the chord Mm with the first diameter Aa, θ β, the angle made b the chord MN with the second diameter Bb, etc Then Euler states without proof that cotθ cotθ cotθ1

6 Figure 11 (We show wh this is true in the notes that follow the translation) It follows that cotθ4 cotθ cotθ1 cotθ5 4 cotθ cotθ1, And in general cot θn ( n 1)cot θ ( n )cotθ1 Section 1 Euler argues that the onl curve in which all parallel lines are diameters is the parabola He does this with the help of the remarkable differential equation of infinite order introduced in section 10 Part II: Curves with diameters passing through a central point Section 14 Having completed the discussion of parallel diameters like those found in parabolas, he turns to diameters that intersect at a common point, as with the ellipse and hperbola In the case of these curves, all straight lines that pass through their center are

7 oblique angled diameters However, are there other curves that have onl one, two, three, etc such diameters? Figure 151 Section 15 Euler now presents the following problem: Find all curves (Fig 151 above) AMm constructed above an axis AC such that the line CF emerging from the point C, making angle ACF with the axis, bisects at E all chords Mm parallel to the tangent line at F Euler now introduces the following notations: t CT and z TM or Tm, m sine of angle ETC, and n cosine of angle ETC, p sine of angle ECT, and q cosine of angle ECT Euler then argues that z, the intersection of all lines TE with the curve, satisfies a quadratic equation z Pz Q, where P and Q are functions of t, and the angle ETC is constant

8 Section 16 Setting x CP and PM Euler gives an eas argument using simple trigonometr that mx + n t and that the famil of desired curves is given b the m equation mp mx + n + x f np mq m where f is an arbitrar function If we let X n x+ and m mp Y + x, Then it np mq follows that this famil of curves is given b W( X, Y ) 0, where W is an arbitrar function of X and Y W( X, Y) α + βx + γy + δx + εxy + ζy + ηx + Section 17 This famil of solutions has onl one guaranteed oblique angled diameter passing through the point C Euler now wishes to find the subset of the above curves that are smmetric about the axis AC In this case AC is an orthogonal diameter Now onl even powers of are allowed Thus the coefficients α, β, γ,, must be determined so that no odd powers of remain in the equation Section 18 First we see that β 0 because no following terms can remove b subtraction n However we can remove odd powers of from the next two terms m γy + δx m p b taking γ np mq and δ, so that our expression becomes n γy + δx mp mpx ( np mq) Y X mq n n Thus Euler lets

9 nnp ( mq) Z nq + mpx mpx Y, m which has onl even powers of Section 19 Now consider the arbitrar function of X and Z: Set W( X, Z) α + βx + γz + δx + εxz + ζz + ηx + W( X, Z ) 0, and this curve has an oblique angled diameter passing through the point C on the x- axis If all the terms in which X appears vanish, then no odd powers of occur and the x- axis is an axis of smmetr for our curve In this case Z is a constant, and Euler writes mpx + nq a These are all conic sections with center at the origin Let b a and get nq b kx, where mp k This means that k tan ETC tan ECT Thus the angle ECT is nq arbitrar, and can be an value An line passing through the origin is an oblique angled diameter for these curves If k 1, the angle TEC is a right angle and the equation is therefore a circle Section 0 Euler now tries to remove odd powers of from the equation W( X, Z) α + βx + γz + δx + εxz + ζz + ηx + 0 He first observes that we must take β δ 0, since the terms involving and x cannot be removed b following terms Next he groups together terms that are homogeneous He starts with X and XZ, which are homogenous of degree, and states (without showing details), that if mp nq, then the terms involving x and disappear He also states

10 5 without proof, that the terms X, XZand XZ, which are homogeneous of degree 5, can be removed if we take np mq or np mq 5+ 5 Section 1 and Euler gives algebraic details of his derivation from the homogenous curves of order discussed in the previous section He lets p q θ and m n 1 θ, and concludes that if we write 1 Z θ + x and V 1 1 θ x θ x 7, and if W denotes an function of Z and V, then the equation W 0 has the desired properties The equation is 0 α + βz + γv + δz + εzv + ζv + ηz + This equation has three diameters, the x-axis which is an orthogonal diameter, and the two lines from the origin making angles θ 1 ± tan with the negative x-axis Section and 4 Now Euler considers the special case of the above equation which is a bz + V 1 1 1 + θ + θ θ, or 7 a b bx x x a 1 1 θ bx + θ x 7 1 b+ θ x These curves are examples of what Euler calls the redundant hperbolas of Newton,

11 Av + Bv + Cv + D v Here 1 v b+ θ x and B C This curve in Figure 41 has four oblique angled 4A diameters all denoted b the same letters CE The point C on the v-axis where these diameters meet has value B v Euler finds other values significant in the graph 6A Figure 41

1 This is Euler s graph to which we have added the v and axies Notice that Euler takes the v axis directed to the left Section 5 Euler has examined curves obtained b starting with X n x+, m 1 Z + θ x and studing the expression of third order α X + βxz He obtained finall the curves V 1 1 x 7 x θ θ, which satisfied his problem In this section he studies the expression of fourth order 4 V αx + βx Z He finds that in this case we must take m p to remove odd powers of He calls p q m n θ and gets ( ) V α θ x Now an function W( Z, V ) 0 will have the x axis as an orthogonal diameter and at least two oblique angled diameters The simplest such 4 4 4 4 curve is given b α θ x

1 Section 6 Using the figure Figure 61 Euler proves that if we have two oblique angled diameters that meet at the point O, Then we have more diameters that meet at O This is a remarkabl simple geometric proof, much like the proof found in section 11 (We give a more detailed proof in the notes) Section 7 Euler gives formulas for important angles in the above figure The calculations are left out, and our detailed derivations of these can be found in the notes Using the notation that all angles are primed and tan x ' x he gets the relations 1 C 1 B + β, and 1+ αb 1 γc + β α B γ + C Part III: An area problem

14 Figure 81 Section 8 Euler now begins a new problem: We seek a curve AMamB that has two orthogonal diameters ACB and ac that are perpendicular to each other Thus the center of this curve is at C Like the ellipse, it must have the following propert: that extending from the center C an ra CM and at the same time another ra Cm parallel to the tangent MT at the point M, the area of the triangle MCm is alwas constant, and is equal to the area of the triangle ACa Section 9 Euler notes that the equation we seek W ( x, ) 0, has, in terms of the variables x and, onl even powers of both variables This follows from the fact that the curve is smmetric about both the x and axies Denote the point M b ( x, ) and the point m ( t, u) If we replace x b t, then is replaced b u In this sense, the two points ( x, ) and ( t, u) are reciprocal Section 0 Euler will tr to find the solution in parametric equations x f ( and g(

15 where z is the parameter He imposes the condition that changing z to z changes x to t and changes to u Thus t f ( and u g( Euler uses P to denote the even part of f and Q to denote the odd part In the same wa R and S denote the even and odd parts of g Thus we have x P( + Q( and R( + S(, and replacing z b z we get t P( Q( and u R( S( Section 1 From the fact that the ra Cm has the same slope as the tangent line at M we get u t d, and therefore dx udx + td 0 An eas calculation shows that area of the triangle MCm is t + ux Since the area is constant, its differential is zero and we get dt + td + udx + xdu 0 Since udx + td is 0, this makes dt + xdu 0 Section Euler denotes the area of triangle MCm b c so that we have t + ux cc Substituting x P + Q, R + S, t P Q and u R S, we get ( P + Q)( R S) + ( P Q)( R + S) cc

16 Euler now introduces V as an odd function of z such that ( P + Q)( R S) cc + V An eas argument now leads to x P + Q, t P Q, cc V P Q and u cc + V P+ Q From these we have after some manipulations ( cc + V )( dp + dq) ( cc V )( dp dq) udx + td dv, ( P + Q) ( P Q) and since this is zero we get (P Q ) dv - V (PdP - QdQ) - cc (PdQ - QdP) 0 Section Dividing the above equation b / ( P Q ) and integrating we get P V Q cc( PdQ QdP) / ( P Q ) So far, we have arbitrar even functions P and Q and odd function R, S and V Euler now begins to specif some of these b tring Q Pz Now the above equation is P V c dz 1 z P(1 z / ) To get an algebraic solution, we must be able to integrate Euler writes differentiating the above equation he gets V Z, and b P P cdz (1 ) z dz + Zzdz Sections 4 and 5 Euler now gets expressions for the main parametric equations x c (1 + dz z dz + Zzdz (1 ) and + c (1 ((1 dz Zd dz ((1 z ) dz + Zzd dz, also

17 t c (1 dz and z dz + Zzdz (1 ) u + + c (1 ((1 dz Zd dz ((1 z ) dz + Zzd dz Here Z is an arbitrar odd function of z Section 6 Euler now considers the simplest case Z x c (1 + ) and αc (1 ; α z α z and gets from which we get αc α x, which is an ellipse Section 7 Now we let parametric equations Z n α z, where n is an odd number, from which we find the x cc(1 + n 1 α z ( n ( n 1) z ), and α n 1 c z z n+ n z (1 ) ( ( 1) ) n ( n 1) z If we let α z Z, we get 1 zz x c (1 + (1 z ) and α(1+ z ) αc (1 z+ z ) + z z z (1 )(1 )(1 )