Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

CHAP. II. Of Parabolical, Hyperbolical and Elliptical Spaces.

Proposition X.

THE(α) 1.1 Parabolick Space (i. e. in Fig. 133. that comprehended under the right line GH and the parabola GEH) is to a cir∣cumscribing Parallelogram GK, as 4 to 6 (or 2 to 3) but to an inscribed ▵ GEH as 4 to 3.

Demonstration.

Suppose FH divided first into two then into four equal parts, and draw parallel to the ax EF the lines ef, ef, &c. dividing also EF into four parts, the first fg will be 3, the second 2, the third 1, by Prop. 34. lib. 1. but as ef is to ge so is ge to he, by Consect. 1. Prop. 4. Therefore he in the diameter EF is = o, in the first ef it is = ¼ (for as ef, 4, to ge, 1, so ge, 1, to he, ¼) in the second ef a portion of he is = 〈 math 〉〈 math 〉, in the third to 〈 math 〉〈 math 〉, and so the portions eh in the trilinear figure EhHK make a series in a duplicate arithmetical progression, viz. 1, 4, 9, 16: After the same manner, if the parts Ff, &c. are bisected, you'l find the portions eh in the external trilinear figure to make this series of numbers ⅛, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, and so onwards. Wherefore since the portions eh or the indivisibles of the trili∣near space circumscribed about the parabola are always in a duplicate arithmetical progression: the sum of them all will be to the sum of as many indivisibles of the parallelogram FK, ••qual to the line KH, i. e. the trilinear space it self to this pa∣ra••lelogram as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Wherefore the semi-parabola ••EhH will be as 2, and the ▵ FEH as 1 ½; therefore the whole parabola as 4, and the whole ▵ GEH as 3, and the whole parallelogram GK as 6. Q E. D.

CONSECTARY I.

IT is evident(α) 1.2 that in the first division, the second line fh (i. e. that drawn from the middle of the base FH) is three such parts whereof FE is 4; for eh is 〈 math 〉〈 math 〉 i. e. 1, therefore fh is 3.

CONSECTARY II.

IT is also evident, that this demonstration will hold of any parabolick segment.

Proposition XI.

THE Elliptical Space(α) 1.3 comprehended by the Ellipsis DAEB (Fig. 127.) is to a circle described on the transverse ax DE, as the Axis Rectus or conjugate diameter AB to the transverse ax DE.

Demonstration.

THis is in the first place evident from the genesis of the ellipse we deduced in Schol. 1. Prop. 7. for in that de∣duction we shewed that FO, i. e. HN was to NI as AB to DE: Which since it is true of all the other indivisibles or or∣dinates HN and IN ad infinitum; it is manifest that the planes themselves constituted of these indivisibles will have the same reason among themselves, as the Axis Rectus AB to the trans∣verse DE. Q. E. D.

CONSECTARY I.

THerefore the quadrature of the ellipse will be evident, if that of the circle be demonstrated.

CONSECTARY II.

SInce a circle described on the least diameter AB will be to one described on the greater diameter DE, as AB to a third proportional by Prop. 35. lib. 1. it follows by vertue of the present Prop. that the ellipse is a mean proportional between the greater and lesser circle, i e. as the ellipse is to the great∣er circle so is the lesser circle to it, viz. as AB to DE.

CONSECTARY III.

HEnce you may have a double method of determining the area of an ellipse. 1. If having found the area of the greater circle, you should infer, as the greater diameter of the ellipsis to the less, so the area of the circle found to the area of the ellipse sought. 2. If having also found the area of the les∣ser circle, you find a mean proportional between that and the area of the greater.

SCHOLIƲM.

WE may also shew the last part of the second Consect. thus, 1. If having described the circle EadbE (Fig. 134.) about the least axis of the ellipse we conceive a regular hexagon to be inscribed, and an ellipse coinciding with one end E of its transverse ax, and with the other or opposite one D to be so elevated, that with the point d it may perpendicu∣larly hang over the circle, and further from all the angles of the figure inscribed in the circle you erect▪ the perpendiculars gG, bB, &c. it is certain that the sides ED and Ed of the triangle DEd will be cut by the parallel planes FGgf, &c. into proportional parts, and that those by reason of the simi∣litude of the ▵ ▵ FDG and fdg, and so also the other rectan∣gles will be among themselves as the intercepted parts of the lines ID and id, CI and ci, and in infinitum, (viz. of how many sides soever the inscribed figure consists:) Wherefore also all the parts of the ellipse taken together will be to all the parts of the circle taken together, i. e. the whole ellipse to the whole

[illustration]

Pag. 201.

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circle as all the parts of the diameter ED or ab, i. e. as DE it self to AB. Q. E D.

CONSECTARY IV.

IT is also evident that both these demonstrations of the pre∣sent Prop. will be also the same in any segments of the el∣lipsis or circle.

Proposition XII.

ANY Hyperbolical space GEHG (Fig. 135) is to any Hyperbolick figure of equal heighth gEhg [whose Latus Rectum and Transversum are equal (as in the circle) and also equal to the Latus Transversum of the former DE, as the Axis Rectus (or conjugate) AB is to the Latus Transversum DE (as in the ellipsis.)

Demonstration.

By the Hypoth. and Prop. 7. and its second Consect. the □ Fg is = □ DFE. Wherefore this □ DFE i e. the □ Fg is to the □ FG as the Latus Transversam to the Latus Rectum of the hyperbola GEHG, by the same seventh Prop. i. e. (by Consect. 2 of the same) as the square of the Latus Transvers. DE to the square of the conjugate AB: Therefore the roots of these squares will be also proportional, viz. Fg to FG as DE to AB; and consequently (since the same is true of any other ordinates ad infinitum) the whole hyperbola gEhg will be to the whole one GEHG as DE to AB. Q. E. D.

CONSECTARY I.

THerefore having found the quadrature of such an hyper∣bola, whose Latus Rectum and Transversum are equal, you may have also the quadrature of any other hyperbola.

CONSECTARY II.

IT is evident that the same demonstration will hold in any other hyperbola's.

Proposition XIII.

ANY Parabolick segments upon the same base, and hyperbo∣lical and elliptical ones described about the same conjugate (one whereof shall be a right one, the other a scalene) and con∣stitued between the same parallels, are equal.

Demonstration.

I. It is evident of Parabola's; for both the right one GEHG, and the scalene one GEHG (Fig. 136. n. 1.) (for the demon∣stration of Prop. 10. will hold in both) is to a ▵ inscribed in them as 4 to 3. But the triangles GEH and GEH are equal, by Consect. 5. Def. 12. or Prop. 28. lib. 1. Therefore the Parabola's also.

Or thus, in the right parabola GEHG every thing is the same as in 1. and 4. Prop. of this Book, viz. EI=eb, EF=ib, the square IK=oecc, the □ FG=oicc. And because therefore in the scalene Parabola also the square FG remains =oicc, make FE=n, and find both the abscissa EI, and the □ answering to it IK.

1. For the abscissa; as FE to EI so FE to EI, per 〈 math 〉〈 math 〉 Consect. 4 Prop. 34 lib. 1.

2. For the □ IK; as FE to EI so □ FG to □ IK, 〈 math 〉〈 math 〉. per Prop. 4. of this.

Therefore the □ IK=□IK and IK=IK, and this in any case ad infinitum: Therefore the one parabola is = to the other. Q. E. D.

II. The business is much after the same way evident of el∣lipses and hyperbolas. For making all things in the ellipsis and right hyperbola (n. 2. and 3. Fig. 136.) as in Prop. 2, 3, 5, 7. viz. the □ IK oecd−eecd in the ellipsis, oecd+eecd in the hyperbola, the □ AB oocd by Consect. 2. Prop. 7. EI=eb, DE=ob, &c. if in oblique ones for the Latus Transversum DE you put n, and seek the Latus Rectum and abscissa EI, you may by means of these also have the square IK, by Prop. 2. and 3.

1. For the Latus Rectum. As n to √oocd so √ oocd to 〈 math 〉〈 math 〉 by Cons. 2.7▪

2. For EI the abscissa. As ob to eb so n to 〈 math 〉〈 math 〉=EI.

3. For the side RS □ deficient or exceeding, from Prop. 2. and 3. As n to 〈 math 〉〈 math 〉 so 〈 math 〉〈 math 〉 to 〈 math 〉〈 math 〉=RS.

Now the abscissa multiplyed by the Lat. Rect.	The abscissa multiplyed by RS.
〈 math 〉〈 math 〉 by 〈 math 〉〈 math 〉 gives □ oecd.	〈 math 〉〈 math 〉 by 〈 math 〉〈 math 〉 gives □ eecd.
The sum of these □ □ oecd + eecd in the hyperb. = □ IK by Prop. 2. evidently = □ IK	The difference of these □ □ oecd−eecd gives in the ellipsis □ IK by Prop. 3. evidently = □ IK.

Wherefore the lines IK and IK, and the whole KL and KL will be equal; and since the same thing is evident after the same way of all other lines of this kind ad infinitum, the elliptical and hyperbolical segments will be so also. Q. E. D.

Notes

• (α) 1.1

  Archim. de Quadratur. Parab. Prop. 17. and 24.
• (α) 1.2

  Archim. Prop. 19. with the Coroll.
• (α) 1.3

  Archim. lib. de Conoid. &c. Prop. 5.