35 If F be the focus of a parabola, and P any point in the tre, the chord of curvature to the point P which pasara through Fis euna! to 4FP.

Let Ps be an Codefinitely small are of the paralela, vin by mi de une forme PHK 'Art 24,; then the 12 may be consitend a comen to both; jos at all, note the lacer to Rad drew paralel to regne Then she gje RP=RHP 19??

datey hear a sincrete vii RP, the trung, Faye sansidered & emang lar PH X

Thus, if P be any point in the curve, then FP+PS=VU=2VC. See the preceding figure.

For (Art. 37.) FV+VS=FU+ US; that is, 2FV+FS= 2US+FS, 2FV2US, and FVUS; and (Art. 37.) FF+ VSFP+SP=VS+US=VU=2VC. Q. E. D.

Cor. 1. Hence, because FV+VS=2VC; by adding VT to both, ST+TF=2CT; and by taking 2TF from this, ST—TF =2CT-2TF=2CF.

Cor. 2. Hence, because (Art. 40.) CVCU, and FV=US (as proved above): CV-FVCU-US, or CFCS.

Cor. 3. Hence, SP=VU-FP=2VC-FP, and in like manner it appears that FP=2VC-SP.

Cor. 4. Hence, because FP+SP=2VC, by taking 2SP from both FP-SP=2 VC—2 SP, or (since SP=2 VC-FP, by cor. 3.)2FP-2VC.

53. The latus rectum is less than 4VF; for BF÷BS=VU (Art. 37.)=2VF+FS (Art. 52.); and since BS is greater than FS, BF must be less than 2VF, and (2BF=) BD less than 41FQ. E. D.

54. A straight line drawn from the focus to the vertex of the minor axis is equal to half the major axis, or FE=VC, See the following figure.

For since FCCS (cor. 1. Art. 52.) and CE common to the two triangles FCE, SCE (and the angles at C right angles (Art. 41.) FE=ES (4. 1.); but (Art. 37.) FE+ES, that is 2FE VU=2VC, FE=VC. Q. E. D.

Cor. 1. And in like manner it may be shewn that FK=KS= ES=EF=VC,. in the triangles FEC, FKC, FK=FE, the angles at Care right angles, and the side FC is common, whence (26.1.) EC=CK.

Cor. 2. Hence EC FEFC2 (47. 1.) VC-FC= (cor. 5. 2.) VC-FC.VC+ FC=VF.FU.

55. If on the major axis as a diameter a circle be described, and the latus rectum be produced to meet the circumference in k, then will Fk=EC. For (14. 2.) Fk2=VF.FU= (cor. 2Art, 54.) EC2, Fk=EC.

56. The latus rectum is a third proportional to the major and minor axes, or VU: EK :: EK: BD.

Because BS=

2VC-BF (cor. 3. Art. 52.). BS2

4VC BF2. 4VC.BF. But BS2 = BF2+ FS2

(47. 1.) ·.· 4VC2 + BF2-4 VC.BF= BF2+ FS2 4VC2 -4VC.BF=FS2

(4. 2.) 4 FC2 ..

(VC2-FC2 by cor.

2. Art. 54.=) EC2=VC.BF (Art. 56.); ·.· (17.6.) VC : EC :: EC: BF; whence (15.5.) VU: EK :: EK: BD. Q. E. D.

Cor. 1. If L (=BD) be the latus rectum, then (since VU · 2VC) L.2VC=EK2 (17. 6.)

Cor. 2. Hence, of the major and minor axes and latus rectum, any two being given, the third may be found.

57. If FP and SP be drawn from the foci, to any point P in the curve, and FP be produced to M, the straight line PT which bisects the exterior angle FPM is a tangent to the ellipse. Make PMPF, join MF, let PT if and join

possible, intersect the curve in p,
Mp, Fp. Then because MP=FP, the

JA &

angle PMF=PFM (5.1.) MPr=FPr by
hypothesis, and Pr common, .* (4. 1.) el mos vigili
Mr Fr, and the angle Mr P=FrP; M
then in the triangles Mpr, Fpr, Mr Fr,
pr common, and the angles at r
equal, (4.1.) Mp=Fp. But (20.1.)
Sp+pM SM, that is > SP+PM, that
is > SP+PF (because PF=PM) that is

are

Sp+pF (because Sp+pF=SP+PF by Art. 37.); since Sp+pM> Sp+pF, if Sp be taken from both pMPF; but it has been shewn that pM=pF, and pFare both equal and unequal to each other, which is absurd; . PT does not

Mp

PA

S

intersect the curve in any other point p; PT is therefore a

tangent at P. Q. E, D.

Cor. 1. It is plain that the nearer the point p be to, the greater will be the angle FpM; and therefore when p coincides with V, the lines Fp, pM will coincide with FV, VT, and the angle FpM will become two right angles; but the tangent at (p which now coincides with) V bisects this angle, the tangent at V is at right angles to the axis VU.

Cor. 2. Hence (prop. A. 6.) ST : TF :: SP: PF.

Cor. 3. Hence, straight lines drawn from the foci to any point in the curve, make equal angles with the tangent at that point, for the angle tPS=MPT (15. 1.)=FPT.

Cor. 4. Hence the triangles FPY, SPt will be similar, and (4.6.) SP: St :: FP: FY.

58. Let P be any point in the ellipse; join FP, SP, then if SG and FG be drawn parallel to these respectively, the point G where they meet will be in the curve.

For since FPSG

is a parallelogram, FG
+ GSSP + FP
(34.1.). G is a point

in the ellipse by Art.
37. Q. E. D.

Cor. Since PG and FS bisect each other in C (part 8. Art. 241. cor.), C is the centre of the ellipse (cor. 1. Art. 52.), and PG a diameter (Art. 42.), all the diameters of the ellipse are bisected by the centre.

59. If Rr be a tangent at G, it will be parallel to Tt.

For since SGr+SGF+FGR=2 right angles (13. and cor. 1. 15. 1.), =SPt+SPF+FPT, and SGF=SPF (34.1.), by taking the latter equals from the former, the remainders SGr +FGR=SPt+FPT, that is, (cor. 3. Art. 57.) 2FGR=2SPt, or FGR SPt; but PGF=GPS (29. 1.); add these equals to the preceding, and FGR+PGF=SPt+ GPS; that is, PGR= GPt, '. (27.1.) Rr is parallel to Tt. Q. E. D.

Cor. Hence, if HD be a conjugate diameter to PG, tangents at D and H will be parallel, and the four tangents Tt, tr, TR, and RT will form a parallelogram circumscribed about the ellipse.

60. If HD be drawn through the centre, parallel to Tt a tangent at P, cutting SP in the point E, then will PE=UC.

Draw FN parallel, and Po perpendicular to HD. Because NF is parallel to tT (30. 1.), and the angles at o right angles, ..the angles oPT, oPt are right angles (29. 1.), or oPT=0Pt, but FPT=SPt (cor. 3. Art. 57.), by taking the latter from the former FPo=NPo, : PNz=PFz (32. 1.), the angles at z the angles at o by 29. 1) right angles, and Pz is common to the triangles PzN, PzF, (26. 1.) PN=PF. And since EC is parallel to NF a side of the triangle SNF, and SC =CF (cor. 1. Art. 52.), . SE=EN (2.6.); ·.· SP+PF (= SN+NP+PF=2 EN+2 NP) = 2 PE. But SP+PF=2 UC (Art. 52.), 2 PE=(SP+PF=) 2UC, and PE UC. Q. ED. 61. If perpendiculars be drawn from the foci to any tangent, and a circle be described on the major axis as a diameter, the points in which the perpendiculars intersect the tangent shall be in the circumference of the circle.

.

Let Ft, ST be drawn Y perpendicular to tT a tangent at P, join SP and produce it to meet Ft produced in Y, and join Ct. Then in the triangles PtF, PtY, the angle tPF=tPY (Art. 57.), the angles at t are right angles, and Pt is common, .' (26.1.) FP

=PY and Ft=tY; also

·:

V

FC=CS (cor. 1. Art. 52.). Ct is parallel to Sy (2. 6.), and the triangles FCt, FSY are similar, FC: Ct :: FS: SY (4.6.). But FCFS, Ct4SY+SP+PY=SP+PF= (Art. 52.) + VU=VC; : since Ct CV, the points t and Vare in the circumference of the circle whose centre is C, and in like manner it may be proved that T is in the circumference. Q. E. D.

U