Thus the equation, with respect to any two conjugate diameters, is similar to that before obtained respecting the two axes.

b'2 a/2

127. If, in the equation y'2 = ( a'2— 212), we make y’— 0,

The curve, therefore, meets the line MM' in two points M and M', equally distant from the centre C; thus, in the ellipse, all the diameters bisect each other at the centre.

it is evident, that if m O be produced, so as to make Om' = 0 m, the point m' will belong to the curve; therefore, each diameter of the ellipse bisects the lines drawn parallel to the tangent which passes through the origin M.

129. We may hence infer; 1. that the tangent, at the extremity N of the diameter NN', is parallel to the diameter MM';

b' a'

2. since y√ α22.

b'

a'

√ (≥ a' + z' ) (¦ a'— z'),

- b',

we may infer, that the ordinates Om of the diameter MM', are
those of the circle, having MM for its diameter, but diminished
or augmented in the ratio of a to b', and making with each other
an angle equal to that of the conjugate diameters. If a' =
these ordinates are precisely equal to those of this same circle.
If we would know in what part of the ellipse the two conjugate
diameters are equal, we have only to find in what part CP = CR,
or CP
CPCR,
CR, or 22 = α2 — z2. Now this equation gives

whence

2

z2 + z2 = 1 a3,

z2 = a2, and z = √ } a2 = }} @ √ },

which may be thus constructed; having described upon the Fig. 61. transverse axis AB (fig. 61), as a diameter, the semicircle ANEB, cut in E by the conjugate axis CD, we bisect the arc AE in N", and letting fall the perpendicular N"P, which cuts the ellipse in M" and M', CM", CM', equal to each other will be the semiconjugate diameters. For, if we designate CP by z,

since the triangle CPN" is right-angled and isosceles, the angle ACN" being 45°, we have

130. If, from the centre C (fig. 63), we draw to TM the Fig. 63. perpendicular CF, the similar triangles TPM, TCF, give

TM: PM:: CT: CF;

whence

PM X СТ

CF=

TM

In like manner, the triangles TPM, CNR, similar on account of the parallel sides, give

TM: TP:: CN: CR;

whence

TM X CR

CN=

TP

TM X TP

consequently, multiplying the above equations, member by mem

ber, we have

CF × CN=

PM X CT X TM × CR PM Х СТ Х CR

or, taking the squares,

TP

substituting these quantities for their equals in the above result,

we have

If now we draw the tangent NT", which produced meets TM in I, CN × CF expresses the surface of the parallelogram CMIN ; anda b, or a × 1b, expresses that of the rectangle of the two semiaxes; therefore, the parallelograms, formed by tangents drawn through the vertices of conjugate diameters, are equal to each other, and equal to the rectangle of the two axes.

131. The similar triangles TPM CRN, give, also, PT: PM:: CR: RN,

but the right-angled triangles CRN, CPM, give

whence

-2

-2

CR+RN= CN, CP + PM = CM,

-2

-2

-2

CR+RN + CP + PM = CN + CM; substituting for the lines in the first member their algebraic values, we have

z) = C + Ci

-2

a2 + b2 = CN + CM;

therefore, the sum of the squares of any two semiconjugate diameters is equal to the sum of the squares of the semiaxes.

132. If, in the equation

-2

CNCR + RN,

-2

we substitute for CR and RN their algebraic values, we shall

Now the similar triangles TPM, MP'T', give, by taking the squares of the homologous sides,

and, multiplying together the members of this equation, and that above, we have

If now we designate the parameter to the diameter MM' by p', we shall have

whence

or

2 CM: 2 CN :: 2 CN: p' (125),

2p' × CM 4 CN,

p' x CM = CN;

therefore, comparing this value of CN with the one above, we

have

TM × MT'p' x CM,

or, in other words,

CM: TM:: MT': P'.

Fig. 64.

If upon TT' (fig. 64), as a diameter, we describe a circle, the circumference will pass through the point C, since the angle TCT' is a right angle. If now we produce CM till it meets the circumference in V, we shall have, by the nature of the circle (Geom. 223),

CM: TM:: MT': MV;

therefore, comparing this proportion with the one above, we obtain

133. From what is here shown we may easily find the axes of an ellipse, and consequently derive a simple method of describing it, when we know only two conjugate diameters MM', NN', and the angle contained by them.

We produce CM, making MV equal to the semiparameter, and from the middle X of CV, we raise a perpendicular XZ, meeting in Z the indefinite line TT', drawn through the point M parallel to NN'; from the point Z, as a centre, and with the distance ZC, as a radius, we describe a circle meeting TT' in two points T, T', through which and the point C, TC, T'C, being drawn, these are the directions of the two axes. We then determine the magnitude of the two axes by letting fall the perpendiculars MP, MP', and taking CA a mean proportional between CT and CP; and CD, a mean proportional between CT' and CP'; for it has been shown (120), that CP: CA: CA: CT; and it is easy to prove, by means of the similar triangles TPM, TCT',

and the known values of TP, PM, and CT, that CT' =

that is, that CP: CD:: CD: CT'.

CD

CP"

Of the Hyperbola.

134. LET us now consider the curve, the property of which is Fig. 65. that the difference of the distances from any point M (fig. 65) to two fixed points F, f, is the same throughout, and equal to a given line a.

We proceed to find, as in the case of the ellipse, an equation, which shalll express the relation between the lines PM perpendicular to Ff, and their distances FP, or AP, from some fixed point F, or A, taken arbitrarily upon the line Ff.

Taking for the origin of the abscissas the point A, determined by applying from the middle C of Ff the line CA equal to ¦ α, I