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Art. VII.—On the Harmonic Conic of Two Given Conics.
[Read before the Philosophical Institute of Canterbury, 4th July, 1908.]
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1. Let two conics be taken—say, S1 ≡ 2l1yz + 2m1zx + 2n1xy = o
S2 ≡ l2yz + 2m2zx + 2n2xy = o.
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These conics are respectively the isogonal transformations of the lines
L1 ≡ l1x + m1y + n1z = 0
L2 ≡ l2x + m2y + n2z = 0
The tangent to S1 at any point P will transform isogonally into a conic touching L2 at the point which is the isogonal conjugate of P.
Let a common tangent to S1 and S2 touch these conies in P1 and P2 respectively, and let its equation be
This transforms isogonally into the conic
which touches L1 and L2 in the points which are the isogonal conjugates of P1 and P2 respectively. Hence, since this conic has double contact with L1 and L2 its equation is of the form
Comparing this with the form
we have p2 = l1l2, q2 = m1m2, r2 = n1n2, and this shows that the four conics which are the isogonal transformations of the common tangents of S1 and S2 touch L1 and L2 along the four lines
These four lines determine on L1 and L2 eight points which are the isogonal conjugates of the points of contact of the four common tangents of S1 and S2. The four points lying on L1 and the four on L2 may be regarded as the intersections of L1 and L2 with a quartic curve.
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The equation of this quartic will be of the form
L1L2(ax2+by2 + 2fyx + 2gzs + 2hxy) + c1c2c3c4=0
where c1≡ √l1l2x + √m1m2Y + √n1n2z
c2≡ √l1l2x - √m1m2Y - √n1n2z
c3≡ -√l1l2x + √m1m2Y - √n1n2z
c4≡ -√l1l2x - √m1m2Y + √n1n2z.
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If we now let a=l1l2, b=m1l2, c=n1n2
2f=-(m1n2 + m2n1), 2g=-(n1l2 + n2l1), 2h=-(l1m2 + l2m)
the quartic reduces to
(m1n2-m1n1)2y2z2 + (n1l2-n2l2-n2l1)2z2x2+(l1m2-l2m1)x2y2+2[(n1l2+n2l2+n2l1)(l1m2+l2m1)x2yz+(l1m2+l2m1)(m1n2+m2n1)y2zx+(m1n2+m2n1)(n1l2+n2l1)z2xy]=0.
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The isogonal transformation of this—on which lie the eight points of contact of the common tangents of S1 and S2—is the conic (m1n2-m1n1)2y2z2+(n1l2-n2l2-n2l1)2z2x2+(l1m2-l2m1)x2y2+2[(n1l2+n2l2+n2l1)(l1m2+l2m1)x2yz+(l1m2+l2m1)(m1n2+m2n1)y2zx+(m1n2+m2n1)(n1l2+n2l1)z2xy]=0.
which is the F conic of the two circumconics S1 and S2.
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The form of the above quartic also shows that on the conic F lie eight points which are the isogonal conjugates of the points in which the four lines √l1l2x±√m1m2y±√n1n2z=0
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intersect the conic F1≡l1l2x2+m1m2y2+n1n2z2-(m1n2+m2n1)yz-(n1l2+n2l1)zx-(l1m2+l2m1)xy=0.
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If (1/l1, 1/m1, 1/n1) and (1/l2, 1/m2, 1/n2) be respectively the co-ordinates of two points o1 and o2, then the conic F1 passes through the six points in which the lines joining the vertices of the triangle of reference to o1 and o2 meet the opposite sides of that triangle; but these six points are the points in which the conics S3=√l1x+√m1y+√n1z=0 S4=√l2x+√m2y+√n2z=0 touch the sides of the triangle of reference.
Hence the conic F1 is the F conic of S3 and S4.
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To obtain the equations of the four common tangents of S1 and S2 let us form the four conies T1≡L1L2-c12=0 T2≡L1L2-c22=0 T3≡L1L2-c32=0 T4≡L1L2-c42=0 and write down their isogonal transformations, which will be the equations of the common tangents.
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Let X0=(√m1n2-√m2n1)2, X1=(√m1n2+√m2n1)2 Y0=(√n1l2-√n2l1)2, Y1=(√n1l2+√n2l1)2 Z0=(√l1m2-√l2m1)2, Z1=(√l1m2+√l2m1)2 then the equations of the common tangents of the two conics S1 and S2 are t1≡X0x+Y0y+Z0z=0 t2≡X0x+Y1y+Z1z=0 t3≡X1x+Y0y+Z1z=0 t4≡X1x+Y1y+Z1z=0
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These tangents constitute four of the group of eight lines represented by the equation (√m1n2±√m2n1)2x+(√n1l2±√n2l1)2y+(√l1m2±√l2m1)2z=0
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The remaining set of four lines may be written p1≡X1x+Y1y+Z1z=0 p2≡X1x+Y0y+Z0z=0 p3≡X0x+Y1y+Z0z=0 p4≡X0x+Y0y+Z1z=0
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Let now the equations of the following conics F1-c12=0, F01-c22=0, F1-c32=0, F1-c42=0 be formed. They are found to be respectively— P1≡X1yz+Y1zx+Z1xy=0 P2≡X1yz+Y0zx+Z0xy=0 P3≡X0yz+Y1zx+Z0xy=0 P4≡X0yz+Y0zx+Z1xy=0 —that is to say, these conics are the isogonal transformations of the four lines p1, p2, p3, p4.
Hence the sixteen points found on F may be regarded as the isogonal conjugates of the sixteen points in which each of the lines c1, c2, c3, c4 is met respectively by the isogonal transformations of the pairs t1, p1; t2, p2; t3, p3; t4, p4.
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The isogonal conjugates of the four points in which c1 is met by the conics T1 and P1 will lie on the conic which is the isogonal transformation of c1, hence the sixteen points on F lie four by four on the four conics It may be easily shown that √l1l2yz±√m1m2zx±√n1n2xy=0. F≡t12p1+4√δ1δ2(l1l2yz+√m1m2zx+√n1n2xy) ≡t2p2-4√δ1δ2(l1l2yz-√m1m2zx+√n1n2xy) ≡t3p3+4√δ1δ2(-l1l2yz+√m1m2zx-√n1n2xy) ≡t4p4+4√δ1δ2(-l1l2yz-√m1m2zx+√n1n2xy) where δ1 and δ2 are the discriminants of S1 and S2.
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It also follows that F≡¼(t1p1+t2p2+t3p3+t4p4)¼Σ(tp).
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The conic F1 passes through the intersection of the conics l1l2x2+m1m2y2+n1n2z2=0 (m1n2+m2n1)yz+(n1l2+n2l1)zx+(l1m2+l2m1)xy=0.
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The former of these is the fourteen-point conic of the system of lines √l1l2x±√m1m2y±√n1n2z=0,
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while the latter is the isogonal transformation of the line L≡(m1n2+m2n1)x+(n1l2+n2l1)t+(l1m2+l2m1)z=0.
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It may be at once shown that F≡L2-δ1δ2S0 where S0 is the conic x2/l1l2+y2/m1m2+z2/n1n2=0.
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It is seen on inspection that t1+p1=t2+p2=t3+p3=t4+p4=2L,σ(t)=σ(p)=4l, and it is easily proved that σ(t2)=σ(p2)=4[F+2δ1δ2S0] t1t2t3t4=f2-4δ1δ2S1S2 = p1p2p3p4-128xyzL.
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It may also be noticed that the lines L1 and L2 are conjugate with respect to all conics inscribed in the standard quadrilateral √l1l2x±√m1m2y±√n1n2z=0 Since t1 touches each of the conics S1 and S2 we have √l1X0+√m1X0+√n1X0=0 √l2X0+√m2X0+√n2X0=0 with similar equations of condition for t2, t3, and t4 hence the coordinates of the four intersections of the conics S3 and S4 are (X0Y0Z0),(X0Y1Z1),(X1Y0Z1),(X1Y1Z0).
The common tangents t2, t3, t4, are the axes of homology of the isogonal conjugates of the intersections of S3 and S4.
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Let the equations of two rectangular hyperbolas S'S” referred to their common self-conjugate triangle be S′≡l′x2+m′y2+n′z2=0 s″=l″x2+m″y2+n″z2=0.
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The equations of their common tangents are x√l′l″(m′n″-m″n′)±y√m′m″(n′l″-n″ll′)±z√n′n″(l′m″-l″m′)=0.
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Let one of these tangents-touch S' and S” at P'(x'y'z') and P″(x″y″z”) respectively, then l′x′=k′x√l′l″(m′n″-m″n′) l″x″k″x√l′l″(m′n″-m″n′), and therefore x′x″=k′k″(m′n″-m″n′); similarly, y′y″=k′k″(n′l″-n″l′) z′z″=k′k″(l′m″-l″m′). Also, since l′+m′+n′=0, l″+m″+n″=0, we have m′n″-m″n′=n′l″-n″l′=l′m″-l″m′, and therefore x′x″=y′y″=z′z″ —that is to say, P' and P” are isogonal conjugates.
Hence the points in which the harmonic conic of S' and S” cuts those conics are isogonal conjugates in pairs with respect to the self-conjugate-triangle of the hyperbolas.