This proposition is equally true when, as in fig. § 6, R is a point in the chord QQ' external to the curve.

This property is seen in the case of the hyperbola in Chap. v. § 11. By Chap. VI. § 19 it appears that as the plane that cuts an hyperbola from the cone moves about the line through A towards the position in which it cuts a parabola, the angle between the asymptotes diminishes and they become more and more nearly parallel to the axis. Hence, as the hyperbola passes into the parabola, the line from any point of the curve parallel to an asymptote becomes parallel to the axis: and the proposition of the present Article takes the place of that previously proved for the hyperbola.

5. The middle points of parallel chords lie on a straight line parallel to the axis.

Let V be the middle point of any chord PQ, draw the

ordinates PN, VO, and the double ordinate QMQ', PR

parallel to the axis.

Then since QQ' is bisected in M,

:

RQ

=

RM + MQ'

=

PN+QM=2VO,

=

and QR. RQ 4AS. PR or QR. VO = 2AS. PR, and VO 2AS :: PR: QR a constant ratio for all parallel chords. Hence VO is invariable for parallel chords, and all their middle points lie on a straight line parallel to the axis.

This line is called the diameter of the chords.

This property corresponds to the fact that the ellipse and hyperbola have the middle points of all parallel chords on a straight line through the centre.

From V the middle point of any chord Qq draw VP parallel to the axis to meet the curve in P. Let PV produced

both ways meet the double ordinates through Q, q (the former produced) in R, r: V will be the middle point of Rr,

and Pr PV + Vr = PV + VR = PR + 2PV.

=

7. The point S we shall see is the focus of the curve, and we may now conveniently consider the properties of the curve with relation to it.

8. Now let AN be the intersection of the plane of the paper with a plane at right angles to it that cuts the cone

QUQ' in the parabola PAP', so that AN is parallel to VQ'.

P

W

K

R

Describe a circle SKL in the plane of the paper touching the generating lines VQ, VQ' in K and L, and AN in S.

Then if we make the circle revolve about its diameter which coincides with the axis, it will generate a sphere which will touch the cone in the circle KRL. Every point of this circle will be equidistant from V. Let a circular section QPQ' through N cut the parabola in P and the plane of the paper in QNQ: the triangle QAN is similar to QVQ':

NAQA and AX= AK, .. NX=QK.

Join SP; also PV cutting the circle KRL in R. PR will QK. And SP, PR tangents drawn to the sphere SKL from the point P are equal: .. SP = PR = QK = NX : hence the distance of any point of the curve from S the distance of the foot of its ordinate from X.

So then if we draw the parabolic section in the plane of the paper and draw through X a line XM at right angles to the axis, and PM at right angles to XM, then we have for any point of the curve SP- NX = PM: the distance of any point P from S= its perpendicular distance from XM. As before, S is the focus and XM the directrix of the parabola.

M

X

A

The eccentricity of the parabola is thus seen to be a ratio of equality. This results also from the fact that the parabola is the limiting form of a semi-ellipse, or of one branch of the hyperbola, when the centre moves to a continually increasing distance from the vertex, whilst the focus approaches a position at a certain definite distance from the

vertex.

In this case AS: AX tends to become a ratio of equality. For in the ellipse and hyperbola CS. CX= CA2; therefore if CS, CX are the distances of an external point C from the points where a line from C through the centre of a circle cuts the circle, CA will be the length of the tangent: and as C moves to a continually increasing distance, the tangent will become more and more nearly parallel to the diameter through C, and CA will tend to become an arithmetic mean between CS, CX. Hence as the semi-ellipse or branch of the hyperbola passes into the parabola, the vertex will assume the position midway between S and X, and the eccentricity will become a ratio of equality.

9. The position of S is identical with that previously assigned to it in proving the relation PN=4AS. AN. For