The distance A'A = __P___

Р 2ep

we shall denote

1-e 1+e 1-e2'

this by 2a; hence the equation becomes

We may suppress the accent if we remember that the origin is at the vertex A', and thus write the equation

y3 = (1 − e3) (2ax — x2)..

II. Suppose the origin at C.

(1).

Since A'C-a, we put x=x+a and substitute this value in (1); thus

We

may suppress the accent if we remember that the origin is now at the centre C, and thus write the equation

y2 = (1 — e2) (a3 — x2)

x=

In (2) suppose x = 0, then y2= (1-e) a2; if then we denote the ordinate CB by b we have b2= (1-e) a2; thus (1) may

-=

be written

161. Since A'S-eOA' and OA'
OA'P

1+e

= (1-e),

162. We may now ascertain the form of the ellipse. Take the equation referred to the centre as origin

For

every value of x less than a there are two values of y, equal in magnitude but of opposite sign. Hence if P be a

point in the curve on one side of the axis of x there is a point P' on the other side of the axis such that P'M= PM. Hence the curve is symmetrical with respect to the axis of x. Values of a greater than a do not give possible values of y; hence, CA being equal to a, the curve does not extend to the right of A.

If we ascribe to x any negative value comprised between 0 and — a, we obtain for y the same pair of values as when we ascribed to x the corresponding positive value between 0 and a. Hence the portion of the curve to the left of YY' is similar to the portion to the right of YY'.

As the equation (1) may be put in the form

we see that the axis of y also divides the curve symmetrically and that the curve does not extend beyond the points B and B', where CB and CB' each = b.

The line E'K' is the directrix; S is the corresponding focus.

=

=

Since the curve is symmetrical with respect to the line YCY', it follows that if we take CH CS and CE CE', and draw EK perpendicular to CE, the point H and the line EK will form respectively a second focus and directrix by means of which the curve might have been generated.

163. The point Cis called the centre of the ellipse because every chord of the ellipse which passes through C is bisected in C. For suppose (h, k) to be a point on the curve, so that the equation

= 1 b2

is satisfied by the values x=h, y=k; then (h, k) is also a point on the curve, because since x=h, y=k, satisfy the above equation, it is obvious that x=-h, y=-k, will also satisfy it. Hence to every point P on the curve there corresponds another point P, in the opposite quadrant, such that

T. C. S.

10

PCP, is a straight line and PC PC. Hence every chord passing through C is bisected in C.

164. We have drawn the curve concave towards the axis of x; the following proposition will justify the figure.

The ordinate of any point of the curve which lies between a vertex and a fixed point of the curve is greater than the corresponding ordinate of the straight line joining that vertex and the fixed point.

Let A' be the vertex, and take it for the origin; let P be the fixed point; x', y its co-ordinates. Then the equation to the ellipse is (Art. 160)

Let x denote any abscissa less than x', then since the

b

ordinate of the curve is √(2ax - x2) or

and that of the straight line is

b

2a

a

(2-1), it is obvious

that the ordinate of the curve is greater than that of the line.

165. AA' and BB' are called axes of the ellipse. The axis AA' which contains the two foci is called the major axis and sometimes the transverse axis; BB' is called the minor axis and sometimes the conjugate axis.

The ratio which the distance of any point in the ellipse from the focus bears to the distance of the same point from the corresponding directrix is called the excentricity of the ellipse. We have denoted it by the symbol e.

To find the latus rectum (see Art. 128) we put x= CH, that is, ae, in equation (1) of Art. 162; thus

=

Since b2= a2- a2e2 ; .. b2 + a2e2 = a2; that is,

CB2+ CH2 = a2;

similarly,

.. BH= a;

BS-a.

166. To express the focal distances of any point of the ellipse in terms of the abscissa of the point.

Let S be one focus, E'K' the corresponding directrix; H the other focus, EK the corresponding directrix. Let P be a point on the ellipse; x, y its co-ordinates, the centre being the origin. Join SP, HP, and draw N'PN parallel to the major axis, and PM perpendicular to it.

Then_SP=ePN' = e (E′′C + CM) = e (~

+ ex.

+ x = a + ex.

Also, HP ePN= e (CE - CM)

=

x= a - ex.