Then, since CH=CS, therefore (Euc. vI., 3) nN=nM=CX. Hence NW is a fixed straight line.

But

LgPH=gHS, from above,

=gNP.

[Euc. I., 29.

Also, the alternate angles GHP, HPG are equal.
Hence the triangles HPN, HPG are similar, and

HP:PN=HG : HP

=SG: SP,

[Euc. VI., 3,

since the angles gPS, gPH stand upon equal circumferences and are equal.

[Euc. III., 27, 28. Therefore, from above, HP bears to PN the constant ratio of SA to AX.

Hence NW has the same properties as the directrix MX. Note. If the result of this proposition be assumed, it may be proved, as in Prop. xv., p. 19, that SP+ PH is constant.

PROP. IV. The normal at any point bisects the angle between the focal distances of the point.

Let the normal at P meet the axis in G.

PROP. V. The tangent at any point is equally inclined to the focal distances of the point.

Let the tangent at P meet the directrix MX in R, PM being perpendicular to MX. Then the circle on PR as diameter passes through the focus S, since PSR is a right angle. [Prop. I., p. 6. It also passes through M for a like reason. Hence the angles SPR, SMR, in the same segment, are equal, and, if

t be any point in RP produced, it may be shown, similarly, that the angles HPt, HNW are equal.

But SX-HW; MX=NW; and X, S are right angles.

Note. The method of Prop. VI., Chap. VI. may be here used.

PROP. VI. The circle which passes through the foci and any point P on the ellipse passes also through the points in which the tangent and normal at P meet the minor axis.

Describe the circle SPH, cutting the minor axis in g, t. Then the equal straight lines gS, gH cut off circumferences which subtend equal angles gPS, gPH. [Euc. III., 27, 28.

Hence Pg bisects the angle SPH, and is therefore the normal at P.

[Prop. IV. Again, gt bisects SH at right angles and is a diameter of the circle. Hence the angle tPg is a right angle, and Pt, being at right angles to the normal, is the tangent at P.

PROP. VII. The tangents at P, Q intersect in T. To prove that

4 STQ=HTP.

Let SP, HQ intersect in O. Produce HP to any point V.

PROP. VIII. To prove that

CA2 - CS2 = CB2 = AS. SA',

where CB is the semi-minor axis and S either focus.

Since CS-CH, therefore

CS+ CB2 = CH2 + CB2.

Therefore SB = HB” (Euc. 1. 47), or SB= HB.

Again, the sum of CS, CA is equal to SA', and their difference to AS. Hence CA- CSAS. SA'. [Euc. II., 5, Cor. CA-CSCB" AS. SA'.

Therefore

=

Note. BC bisects the angle SBH and is normal at B.

[Prop. IV.

Also MBN, drawn parallel to the axis to meet the directrices in M, N, is at right angles to the normal BC and touches the curve at B.

Hence BSM is a right angle (Prop. I., p. 6); and since the angles SBC, SMB are equal, each of them being complementary to SBM, therefore the right-angled triangles SBC, SMB are similar.

Now SB-CA. Also (i) B is a point on the curve; and (ii) BM=CX. Therefore

CS: CA

=

SA: AX

(i),

(ii),

PROP. IX. The foot of the perpendicular drawn from either focus to the tangent at any point lies on the auxiliary circle.

Let CY, drawn parallel to HP, meet the tangent PY in Y and SP in O. Produce YP to Z.

Then, because CY is parallel to HP, and CS=HS, therefore COHP (Euc. IV., 2), and OS=SP=OP.

Therefore

is the centre of the circle round SPY and the

angle SYP, in a semi-circle, is a right angle.

Therefore Y lies on the auxiliary circle; and it has been shewn that SYP is a right angle.

Similarly, if HZ be drawn to meet the tangent YPZ at right angles, then Z lies on the auxiliary circle.

COR. Complete the parallelogram PYCk by drawing the diameter parallel to the tangent at P or perpendicular to the normal PF.

where SY, HZ are the focal perpendiculars upon the tangent at any point P.

Describe the auxiliary circle, passing through Y, Z (Prop. IX.), and let ZH meet YC in V.

Then YV is a diameter of the circle, since YZV is a right angle. [Construction.