PROPOSITION 1. THEOREM.

In ellipses (as APB, fig. 1), the sum, and in hyperbolas (as AQ, BP, fig. 2), the difference, of the distances (EP, PF) of any point (P) of the section, or opposite sections, from the focuses. (E and F), is equal to the principal axis (AB); and either axis (AB or MN) is bisected in the centre (C).

Part 1. In the ellipse (see fig. 1), the sum of AE and AF, or of twice AE and EF, is equal to the sum of FB and EB (Def. 2. 1 Sup.), or of twice FB and EF; taking away EF which is common, twice AE is equal to twice FB, and so AE to FB (Ax. 7. 1 Eu.); therefore AF and AE together are equal AF and FB together, or AB; but EP and PF together are equal to AF and AE together (Def. 2. 1 Sup.), therefore EP and PF together, are equal to AB.

Part 2. In like manner, in the hyperbola (see fig. 2), the difference of AF and AE, or of EF and twice AE, is equal to the difference of EB and FB (Def. 4. 1 Sup.), or of EF and twice FB, wherefore, taking from EF that difference, the remainder is equal to either twice AE or twice FB, which are therefore equal to each other (Ax. 3. 1 Eu.), and so AE is equal to FB (Ax. 7. 1 Eu.); therefore the difference of AF and ÃE, is equal to the difference of AF and FB, or AB; but the difference of EP and PF is equal to the difference of AF and AE (Def. 4. 1 Sup.); therefore the difference of EP and PF is equal to AB.

Part 3. And, AE having been proved equal to BF, and EC being equal to CF (Def. 3 and 5. 1 Sup.), AC is equal to CB (Ax. 2 and 3. 1 Eu.), and so AB is bisected in C.

And in fig. 1, EM and MF being joined, the triangles ECM and FCM, having CE and CF equal (Def. 3. 1 Sup.), CM common, and the angles at C right (by the same), ME and MF are equal (4. 1 Eu.); whence, EM and MF together being

equal to AB (part 1, of this), either of them, as FM, is equal to its balf CB; in like manner FN may be proved equal to CB; therefore FM and FN are equal, and of course the angles FMN and FNM (5. 1 Eu.), whence, the triangles MCF and NCF having also the right angles at C equal, MC is equal to CN (26 1. Eu.), and so MN is bisected in C.

In fig. 2, MB and BN, being drawn, are equal (Def. 5. 1 Sup.), and therefore the angles BMN and BNM (5. 1 Eu.); whence, the triangles MCB and NCB, having also the right angles at C equal, MC is equal to CN (26.1 Eu.), and so MN is bisected in C.

Cor. "The distance of a vertex (M, see fig. 1), of the second "axis (MN) of an ellipse from either focus (E or F), is equal "to the principal semiaxis." It having been proved, in the demonstration of part 3 of this proposition, that either EM or MF is equal to CB.

PROP. II. THEOR.

The square of the second semiaxis (CM, see fig. to prec. prop.), of an ellipse or hyperbola (BP), is equal to the difference of the squares of the principal semiaxis (CB), and the eccentricity (CF), or to the rectangle under the distances of either focus (as F), from the principal vertices (A and B), or of either principal vertex (as B), from the focuses (E and F).

In the ellipse (see fig. 1), draw MF, which is equal to CB [Cor. 1. 1 Sup.]; whence, in the triangle CFM, right-angled at C, the square of CM is equal to the difference of the squares of MF and CF [47. 1 Eu.], or of CB and CF, or, which is equal [Schol. 6. 2 Eu.], to the rectangle AFB or EBF.

In the hyperbola (see fig. 2), draw MB, which is equal to CF [Def. 5. 1. Sup.]; whence, in the triangle CBM, right angled at C, the square of CM is equal to the difference of the squares of MB and CB [47. 1 Eu.], or of CF and CB, or, which is equal [Schol. 6. 2 Eu, to the rectangle AFB or EBF.

Cor 1. Hence, if, in ellipses, there be taken two points (E and F, see fig. 1), in the principal axis (AB), the rectangle under the distances of each of which from the principal vertices (A and B, namely, the rectangle AEB or AFB), is equal to the square of the second semiaxis (CM), which may be done by cor. 2. 5 and 6. 2 Eu. these points are the focuses of the ellipse. Cor. 2. In like manner, if, in hyperbolas, there be taken two points (E and F, see fig. 2), in the principal axis (AB) pro

*

duced both ways, the rectangle under the distances of each of which from the principal vertices (A and B, namely, the rectangle AEB or AFB), is equal to the square of the second semiaxis (CM), which may be done by cor. 3. 5 and 6. 2 Eu. these points are the focuses of the hyperbola or opposite hyperbolas.

PROP. III. THEOR.

The sum of the distances GE and GF, see fig. 1), of any point (G) without an ellipse (APB), from the focuses (E and F), is greater, of any point (H) within it, less, than the principal axis (AB).

And the difference of the distances (GE and GF, see fig. 2), of any point (G) without opposite hyperbolas (AO and BP), from the focuses (E and F), is less, of any point (H), within one of them (BP), greater, than the principal axis (AB).

And the distance (GF, see fig. 3), of any point (G) without a parabola (KP, from the focus (F), is greater, of any point *(H) within it, less, than the distance of the same point from the directrix (DO).

Part 1, fig 1. Let FG, or FH produced, meet the ellipse in P; and EG and GF together, are greater, and EH and HF together, less, than EP and PF [21. 1 Eu.], or, which is equal [1. 1 Sup], AB.

Figl

A

Part 2, fig. 2. Let GF meet the hyperbola, BP in P, and because EG is less than EP and PG together [20. 1 Eu], the excess of Fig.2. EG above GF, is less than the excess of EP and PG together above GF [Ax. 5. 1 Eu.], or, of EP above PF, or [1. 1 Sup.], AB. And the difference of EH and HF is greater than AB.

For, having drawn HE meeting the hyperbola BP in K, and joined KF; because HF is less than HK and KF together [20. 1 Eu.], the excess of EH above HF, is greater than the excess of EH above HK and KF together, or of EK above KF, or [1. 1 Sup.], AB. Part 3, fig. 3. Join GF and HF, and draw GL and HD perpendicu

E

F

F

Fig.31

F

lar to DO; and, because G is without the parabola and F within it, GF meets the parabola, as in P, and, for the same reason, HD meets it, as in K; draw PO perpendicular to DO, and join KF and GO; and PF and PO being equal [Def. 8. 1 Sup.], GF is equal to GP and PO together [Ax. 2. 1 Eu], and therefore, GP and PO together being greater than GO [20. 1 Eu.], and GO than GL[Cor. 1. 19. 1 Eu.], GF is greater than GL. And HF is less than HK, KF together [20. 1 Eu.], or, KB being equal to KF [Def. 8. 1 Sup.], than HD.

PROP. IV. THEOR.

If the sum of the distances of any point from the focuses (E and F, see fig. 1 above), of an ellipse, be equal to the principal axis (AB), that point is in the ellipse; if that sum be greater, it is without, if less, within the same.

If the difference of the distances of any point from the focuses (E and F, see fig. 2 above), of a hyperbola, be equal to the principal axis, that point is in one of the opposite hyperbolas; if that difference be less, it is, without both, if greater, within one of them.

And if the distance of any point from the focus (F, see fig. 3 above), of a parabola, be equal to its distance from the directrix (DO), that point is in the parabola; if the distance from the focus be greater, the point is without, if less, within it.

Part 1. Fig. 1. If the sum of EP and PF be equal to AB, P is in the ellipse; for, if P were without the ellipse, that sum would be greater, if within it, less, than AB [3. 1 Sup.], contrary to the supposition.

If the sum of EG and GF be greater than AB, G is without the ellipse; for, if G were in the ellipse, that sum would be equal to, if within it, less than AB [1 and 3. 1 up], contra hyp.

And if the sum of EH and HF be less than AB, t is within the ellipse; for, if H were in the ellipse, that sum would be equal to, if without it, greater than AB (1 and 3. 1 Sup.), contra hyp. Part 2, fig. 2. If the difference of EP and PF be equal to AB, P is one of the opposite hyperbolas; for, if P were without both of them, that difference would be less, if within one of them, greater, than AB (3, 1 Sup.), contra hyp.

If the difference of EG and GF be less than AB, G is without both of the opposite hyperbolas; for, if G were in

one of these hyperbolas, that difference would be equal to, if within one of them, greater than, AB (1 and 3. 1 Sup.), contra hyp.

And if the difference of EH and HF be greater than AB, H is within one of the opposite hyperbolas; for, if I were in one of them, that difference would be equal to, if without both of them, less than, AB (1 and 3. 1 Sup.), contra hyp.

Part 3, fig. 3. If PF drawn to the focus be equal to PO drawn perpendicularly to the directrix, P is in the parabola ; for, if P were without the parabola, PF would be greater, if within it, less than PO (3. 1 Sup.), contra hyp.

If GF drawn to the focus be greater than GL drawn perpendicularly to the directrix, G is without the parabola; for, if G were in the parabola, GF would be equal to, if within it, less than, GL (1 and 3. 1 Sup.), contra hyp.

And if HF drawn to the focus be less than HD drawn perpendicularly to the directrix, H is within the parabola; for, if H were in the parabola, HF would be equal to, if without it, greater than, HD (1 and 3. 1 Sup.) contra hyp.

PROP. V. THEOR.

Any diameter of an ellipse or hyperbola is bisected in the centre.

If the diameter be an axis, the Fig.1 proposition is demonstrated in 1. 1 Sup.

But if it be any other diameter, A as QP, (see fig. 1 and 2), C being the centre, QP is bisected in C.

For if QC and CP be not equal, let one of them, as CP, if possible, be greater than the other QC, and take on CP, a part CR equal to CQ, and to the focuses E, F, draw PE, PF, QE, QF, RE and RF, and join EF, producing it, in fig. 1, both ways, to the principal vertices A and

B.

The triangles ECR, QCF (see both fig.), having the sides EC, CR and angle ECR, severally equal to FC, CQ and the angle FCQ, the right

E

B

Fig. 2.

E

H