the rectangle DHG, or, which is equal (5. 2 Eu.), the square of CH, is equal to the excess of the square of CG above that of CT, or (5. 2 Eu.), the rectangle DTG.

And, in the case of fig. 2, the demonstration of the equality of the square of CT to the rectangle DHG, in the preceding paragraph, is applicable to this case and its figure, without variation. Adding to these equals the square of CG, the rectangle DHG with the square of CG, or which is equal (6. 2 Eu.), the square of CH, is equal to the squares of CG and CT together.

Cor. Hence in the ellipse, the squares of the segments (CH and CT), of the diameter (DG), to which ordinates are drawn from the vertices of two conjugate diameters, between the centre and ordinates, are together equal to the square of the semidiameter (CG) to which they are so drawn. For since the square of CH is, by this proposition, equal to the rectangle DTG, the square of CH within the square of CT, is equal to the rectangle DTG and the square of CT, or which is equal (5. 2 Eu.), the square of CG.

PROP. LI. THEOR.

If from a vertex of each of two conjugate diameters (CP and CO, see fig. 2 to the preceding proposition), of a hyperbola, ordinates (PH and OS) be drawn to two other conjugate diameters (CG and CR); the segments (CH and CS) of the diameters, to which the ordinates are drawn, between the centre and the ordinates, are directly, and the ordinates themselves (PH and OS) inversely, as the semidiameters (CG and CR) to which they are drawn.

For the sum of the squares of CG and CT, or which is equal (50. 1 Sup.), the square of CH, is to the square of OT or CS, as the square of CG is to the square of CR (40. 1 Sup.), therefore CH is to CS, as CG is to CR (22. 6 Eu).

And the rectangle DHG, or which is equal (50. 1 Sup), the square of CT or OS, is to the square of PH, as the square of CG to the square of CR, (40. 1 Sup.), therefore OS is to PH, as CG to CR (22. 6 Eu).

PROP. LII. THEOR.

If one extreme (D), of a right line (DP), equal to the principal semiaxis (CB) of an ellipse, be in the second axis (MN", and the segment thereof (DG) between the axes, be equal to the difference of the semiaxes (CB and CM); its other extreme (P) is in the perimeter of the ellipse.

and CH, or, (5. 2 Eu.), the rectangle

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AHB is to the square of DP or CB; therefore the point P is in the perimeter of the ellipse Cor. 3. 40. 1 Sup.).

Cor. Hence, if two unequal right lines AB and MN, of which AB is the greater, bisect each other at right angles in C, and a right line DG be placed between AB and MN equal to the difference of their halves CB and CM, and on DG produced, GP be taken equal to CM, and the right line DGP be so moved through the four right angles, that the point D may be always in the right line MN, and G in AB; the point P would describe an ellipse, whose transverse axis is AB, and second axis MN. And hence an ellipse is described, by means of an instrument, called an Elliptick Compass, as to any one viewing its structure, may hence easily appear.

PROP. LIII. THEOR.

A parallelogram (LSIZ), described about two conjugate diameters (QP and RO), of an ellipse or hyperbola, by drawing through their vertices, four right lines, touching the ellipse or conjugate hyperbolas, is equal to the rectangle under the axes (AB and MN).

Let C be the centre, and SPL produced if necessary, meet AB in K, and MN in X, let PH and PT be drawn, at right angles to the axes, and PD and CG at right angles to SPL.

Then the right angled triangles (see fig. 1), PHD and CGX, being, because of the parallelism of the sides forming the angles DPH and GCX, equiangular (Cor. 3. 9. 1 Sup.), PH is to PD, as CG is to CX (4. 6 Eu.), therefore the rectangle under PD and GC is equal to the rectangle under PH and CX (16. 6 Eu.), or to the rectangle TCX, or, which is equal (44. 1 Sup. and 17. 6 Eu.), the square of CM; and therefore CG, CM and PD are continually proportional (17. 6 Eu.); therefore the square CG is to the square of CM as CG is to PD (Cor. 2. 20. 6 Eu.), or, because of the equiangular triangles CGK and DPK, as GK is to PK, or, which is equal (1. 6 Eu.), as the rectangle GKP, or, because of the equiangular triangles CGK and PHK, the rectangle CKH (4 and 16. 6 Eu.), or (46. 1 Sup.), AKB to the square of PK, or, which is equal (41. 1 Sup.), as the square of CB to the square of CO; therefore CG is to CM as CB is to CO (22. 6 Eu.), and so the rectangle under CO and CG, or, which is equal (35. 1 Eu.), the parallelogram COLP, is equal to the rectangle under CB and CM (16. 6 Eu.); whence the parallelogram LSIZ being fourfold the parallelogram COLP (23. 6 Eu.), and the rectangle under AB and MN fourfold that under CB and CM (by the same), the parallelogram LSIZ is equal to the rectangle under AB and MN.

The reasoning of the preceding paragraph, applies to the case of fig. 2 without variation.

Cor. 1. All parallelograms, described about conjugate diameters, of a given ellipse or hyperbola, by drawing tangents

through their vertices, are equal to each other, being each of them (by this prop.), equal to the same rectangle.

Cor. 2. All parallelograms, formed by joining the vertices of conjugate diameters of a given ellipse or hyperbola, aro equal; being halves of the parallelograms treated of in the preceding corollary.

PROP. LIV. THEOR.

The sum of the squares of any two conjugate diameters of an ellipse, is equal to the sum of the squares of the axes. And if the angles contained by the asymptotes of a hyperbola be right, any two conjugate diameters are equal. But if the angles contained by the asymtotes be not right, any two conjugate diameters are unequal; and the difference of the squares of any two conjugate diameters, is equal to the difference of the squares of the axes.

Part 1. Let CB and CM be semiaxes of an ellipse, AB and MN being the axes, and CP and CO two other conjugate semidiameters, let PH and OT be perpendicular to A CB, and PD and OG to MN.

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Q

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Because the square of CB is equal to the squares of CH and CT together (Cor. 50. 1 Sup.), and the square o CM to the squares of CD and CG together by the same), or to those of PH and OT; the squares of CB and CM together are equal to the four squares of CH, CT, PH and OT, to which the squares of CP and Co are also equal (47. 1 Eu.), therefore the squares of CP and CO together are equal to those of CB and CM together; but the squares of the diameters QP and RO together are fourfold the squares of CP and CO together (Cor. 4. 2 Eu.), and the squares of AB and MN together are fourfold the squares of CB and CM together (by the same); therefore the squares of QP and RO together are equal to the squares of AB and MN together (Ax. 1. 5 Eu).

Part 2. Let CD and CK be the asymptotes of a hyperbola, whose centre is C; let CP be any semidiameter drawn to the hyperbola BP, and through its vertex P, let a tangent DPG be drawn (48. 1 Sup.), meeting the asymptotes in D and G, I'D or PG is equal to the semidiameter conjugate to CP (Cor. 2. 37. 1 Sup). But if the angle DCG be right, a circle described from the centre P about the diameter DG would pass through C (Cor. 31. 3 Eu.), and of course the semidiameter CP would be equal to PD or PG, or, by Cor. 2. 37. 1 Sup, to the semidiameter conjugate to it, and so the diameter drawn from P would be equal to its conjugate (Ax. 6. 1 Eu).

In this case, the hyperbola is said to be, Equilateral or Rightangled (see Def. 20. 1 Sup).

Part 3. Det now the asymptotes CD and CK contain an acute angle, and let CP be any semidiameter; draw the transverse semiaxis CB (35. 1 Sup.), through P and B draw the tangents DG and HK, meeting the asymtotes in D and G, H and K; from P, D, B and H, let fall on the asymptote CK, the perpendiculars PQ, DO, BR and HL.

Because the angle DCG is acute (Hyp.), it falls without a semicircle described about DG as a diameter, for otherwise the angle GCD would be right or obtuse (31. 3 and 16. 1 u.), contrary to the supposition, therefore the semidiameter CP of the hyperbola is greater than PD or PG, and therefore than its conjugate semidiameter (Cor. 2. 37 and Def. 14. 1 Sup).

And because DG and HK are bisected in P and B (37. 1 Sup.), OG and LK are, because of the parallels, bisected in Q and R (2. 6 Eu.); and since the rectangle GCD is equal to the rectangle KCH (Cor. 4. 38. 1 Sup.), CK is to CG, as CD is to CH (16. 6 Eu.), or, because of the parallels, as CO is to CL (2. 6 Eu.); therefore the rectangle GCO is equal to the rectangle KCL (16. 6 Eu.); but the rectangle GCO is equal to the difference of the squares of CQ and QG (6. 2 Eu.) or (Cor. 1 and Schol. 6. 2 Eu.), to the difference of the squares of CP and PG; and for the same reason the rectangle KCL is equal to the difference of the squares of CR and RK, or of CB and BK ; therefore the difference of the squares of CP and PG, is equal to the difference of the squares of CB and BK ; but PG and BK are equal to the semidiameters which are conjugate to the semidiameters CP and CB (Cor. 2. 37. and Def. 14. 1 Sup.), there