to 12. 2 Sup.; and the diameters LM and NP pass by the axis of the cone (Cor. 21. 2 Sup.); therefore the plain ANLBCMP passes by the axis of the cone; let the right line DF be the intersection of this plain with the plain of the circle DEFG; since therefore DF passes through the points R and S, and therefore bisects the parallels EG and OQ, it is a diameter of the circle DEFG, for if any other fight line, bisecting one of them not passing through the centre, were a diameter, it would bisect it perpendicularly (3. 3 Eu.), and would therefore, because of the parallels, be perpendicular to the other (29. 1 Eu.), and therefore would bisect that other (3. 3 Eu.), which would of course be bisected in two points, which is absurd; and that DF is a diameter appears also from Cor. 2. 32. 1 Sup,, and it is perpendicular to the right lines EG and OQ (3. 3 Eu.); since therefore EG is perpendicular to the right lines LRM and DRF, it is perpendicular to the plain ABC passing by them (2. 2 Sup.), and therefore the plains DEFG and LEMG and of course the plain of the base, are perpendicular to the plain ABC passing by the axis (9. 2 Sup.); but because of the circles LEM and DEF, each of the rectangles LRM and DRF are equal to the square of ER (3 and 35. 3 Eu.), and therefore to each other, therefore DR s to LR, as RM is to RF (16. 6 Eu.), and the angles DRL and FRM are equal (15. 1 Eu.), therefore the triangles DRL and MRF are equiangular (6. 6 Eu.), and the angle MFR is equal to DLR, or its equal (29. 1 Eu.), ABC; therefore the section DEFG is a subcontrary section (Schol. 22. 2 Sup).

Cor. Hence the section made by a plain with a conical surface, which is neither a subcontrary section, nor made by a plain, which is parallel to the base, is not the circumference of a circle.

PROP. XXIV. THEOR.

If a cone be cut by a plain neither passing through its vertex, nor parallel to its base, nor placed subcontrarily; the section is an ellipse, hyperbola or parabola; the intersection of that plain with a plain passing by the axis of the cone, and a right line drawn through the centre of its base, perpendicular to the common section of that plain with the base of the cone, being a diameter of the section; the section being an ellipse; when this diameter meets the conical surface of the cone twice; a hyperbola, when it meets the opposite conical surfaces; and a parabola, when being parallel to a right line drawn from the vertex of the cone to the circumference of its base, it meets only one of the conical surfaces, and but once.

Let a cone ABC, see fig. 1, 2 and 3, be cut by a plain PEF, neither passing through its vertex, nor parallel to its base, nor placed subcontrarily ;let HK be the common section of this plain with the plain of the base, D the centre of the base, DG a right line, drawn through D perpendicular to HK, ABC the triangu lar section of the cone with the plain passing by the axis and DG, and EG the intersection of the plains EF and ABC; the section PEF is an ellipse, hyperbola or parabola, of which EG is a diameter; the section being an ellipse, when the diameter EG meets the conical surface of the cone ABC twice, as in E and L in fig. 1; a hyperbola, when it meets the opposite conical surfaces, as in E and L in fig. 2; and a parabola, when the diameter EG, being parallel to a right line AC, drawn from the vertex A of the cone, to the circumference of its base, meets only one of the conical surfaces, as in E in fig. 3, and but once.

Let first EG meet the conical surface of the cone ABC in E and L, and let there be taken in the section any point P, and draw PO parallel to HK meeting EL in O, and thro' O draw MN parallel to BC; the plain passing by MN and PO is parallel to that passing by BC and HK (14. 2 Sup.); therefore the section by the plain POM is a circle whose diameter is MN (21. 2 and Cor. 21. 2 Sup.), and because the angle BCH is a right angle (Hyp.), B MOP is a right angle (12. 2 Sup.),

Fig. 1.

M

A

therefore the square of PO is equal to the rectangle MON (3 and 35. 3 Eu). In like manner, if any other point T be taken in the section PEL, and TQ be drawn to EL parallel to HK or PO, and through Q, RS be drawn parallel to BC, it may be shewn, that the square of TQ is equal to the rectangle RQS; therefore the square of PO is to the square of TQ, as the rectangle EOL is to the rectangle EQL (Cor. 1. 7. 5 Eu.); but, because of the equiangular triangles EMO and ERQ, LON and LQS, MO is to RQ, as EO is to EQ, and ON is to QS, as OL to QL (4. 6 and 16. 5 Eu.), therefore the ratios of the rectangle MON to the rectangle RQS, and of the rectangle EOL to the rectangle EQL, being compounded of these equal ratios (23. 6 Eu.), are equal (22. 5 Eu.); therefore the square of PO is to the square of TQ, as the rectangle EOL is to the rectangle EQL (11. 5 Eu). Let an ellipse be described with the diameter EL, its vertices being E, and L, and the ordinate PO (85. 1 Sup.), and because the square of TQ is to the square of PO, as the rectangle EQL is the rectangle EOL, the point T is in the ellipse (Cor. 3. 40. 1 Sup.): in like manner it may be shewn that any other point in the section PEL is in the ellipse; and since the section, being neither a subcontrary one, nor parallel to the base of the cone (Hyp.), is not a circle (Cor. 23. 2 Sup.), it is an ellipse, of which EL is a diameter.

Secondly Let EG produced meet the opposite conical surface in L, let H and K be the points in which HK meets the circumference of the base of the cone, from any point P in the section HEK, draw PO parallel to HK, meeting EG in O, and through O draw MN parallel lo BC; as before, the plain by PO and MO is parallel to that by HG and BC, or to the base of the cone, and the angle POM equal to HGB, and therefore a right angle, therefore the square of PO is equal to the rectangle MON (3 and 35.

3 Eu.); in like manner it may be proved, that the square of HG is equal to the rectangle BGC; therefore the square of PO is to the square of HG, as the rectangle MON is to the rectangle BGC Cor. 1. 7. 5 Eu.); but, because of the equiangular triangles

Fig. 2.

M

L

N

P

K

At

MOE and BGE, LON and LGC, MO is to BG, as EO is to EG, and ON to GC, as LO to LG; therefore the ratios of the rectangle MON to BGC and of EOL to EGL, being compounded of these equal ratios (23. 6 Eu.), are equal (22. 5 Eu); therefore the square of PO is to the square of HG, as the rectangle EOL is to the rectangle EGL (11.5 Eu). Let a hyperbola be described with the diameter EL, its vertices being E and L. and the ordinate HG (85. 1 Sup.), and, because the square of PO is to the square of HG, as the rectangle EOL is to the rectangle EGL, the point P is in the hyperbola (Cor. 3. 40. 1); in like manner it may be shewn, that any other point in the section HEK is in the hyperbola, and therefore that section is a hyperbola, of which EL is a diameter.

Thirdly. Let EG be parallel to a right line AC drawn from the vertex A of the cone to the circumference of its base, let H and K be the points in which HK meets that circumference, from any point P in the section HEK draw PÓ parallel to HK, meeting EG in O, and through O draw MN parallel to BC; as before, the plain by PO and MN is parallel to that by HG and BC, or to the base of the cone, and the angle POM equal to HGB, and therefore a right angle, therefore the square of PO B is equal to the rectangle MON (3 and 35. 3 Eu). In like manner it may be

Fig. 3.

M

P

H

E

F

K

G

shewn, that the square of HG is equal to the rectangle BGC; therefore the square of PO is to the square of HG, as the rectangle MON is to the rectangle BGC (Cor, 1. 7. 5 Eu.), or, because of the equals ON and GC (34. 1 Eu.), as MO is to BG (1. 6 Eu.), or, which is, because of the equiangular triangles MOE and BGE, equal (4. 6 and 16. 5 Eu.), as EO is to EG. Let a parabola be described with the diameter EG, its vertex being È, and ordinate HG (85. 1 Sup.), and because the square of PO is to the square of HG, as EO is to EG, the point P is in the parabola (Cor. 3. 40. 1 Sup.); in like manner it may be shewn, that any other point in the section HEK is in the parabola, and therefore that section is a parabola, of which EG is a diameter.

Scholium. From the five preceding propositions it appears, that the figures formed by the intersection of a plain with a conical surface, are the same, as those defined in Def. 1. 1 Sup.

397

ELEMENTS OF SPHERICAL TRIGONOMETRY.

NOTE.-In subsequent citations, Sph. Tr. denotes, Spherical Trigonometry.

DEFINITIONS.

1. A great circle of a sphere, is one, whose plain passes through the centre of the sphere.

2. The pole of a circle of a sphere, is a point in the surface of the sphere, from which all right lines drawn to the circumference of the circle are equal.

3. A spherical angle, is an angle on the surface of a sphere, contained by the arches of two great circles which meet each other; and is the same, as that which the plains of these circles make with each other; being the angle contained by two right lines, drawn in the plains of these circles, perpendicular to the line of common section, from the same point therein.

4. A spherical triangle, is a figure on the surface of a sphere, comprehended by the arches of three great circles.

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