we shall have fm+mF-FM+Mfor fM-fm=Mr=Fm-FM =mg; therefore the right-angled triangles mMg, mMr, are equal and similar, and consequently the angle gmM, or FmT, or FMT* =mMr=LMT. Therefore if we prolong the radius vector fM, the line MT which bisects the angle LMT, will be the tangent required.

The angle LMT=OMf=FMT, therefore any rays proceeding from a luminous focus F, would on meeting the ellipse AM be reflected to the other focus f.

32. If we draw the normal MN, the angle fMN will be equal to the angle NMF (because ∠OMF+ZfMN=∠ TMF + ∠ NMF, and we have just shewn that / OMF, or its equal LMT=TMF) therefore we shall have fM: FM::fN: FN (Euclid), or by composition

or since c2= a2-b2, by substitution we have FN=c-z+

62 z

b2 z

az

Consequently FN + z-c=PN= P. This is the expression

a2-2a

for the subnormal in the ellipse, when the origin of the abscissæ is at the center. If it were at the vertex, call AP, x as before, since

(When the origin of the abscissæ is at the vertex PT

aa

Consequently CT=PT+CP=, which gives this proportion

z

CP:CA::CA: CT

by which it is easy to determine the point T, through which the tangent MT passes.

• Because the arc Mm being indefinitely small, the angles FmT, FMT may be considered as equal.

The expression for the tangent MT may be found by means of the right-angled triangle PMT

of the diameter CN, and the parts CP are the abscisse. Lastly, the parameter or latus rectum of any diameter, is a third proportional to that diameter and its conjugate.

34. From the extremities D and N draw the two ordinates NQ, DI to the transverse axis Aa, call CQ=z, DI=u; because of the similar triangles DIC, NQT, we shall have

bb

aa

bz

Hence u =

a

and therefore CQ (2) : DI (u)::a:6; in the

very same manner we shall find CI: NQ::a: b. Consequently CQ: DI :: CI: NQ; therefore the triangles DIC CNQ, are equal in surface.

Hence Is, DI*b* CQ

-bb-NQ2, or DI2+NQ-66

α

IIo, CI =

NQ=aa-CQ, or CI+CQ2= a2

III°, a2+b2 CI+CQ+NQ+DI-CD+CN2; that is, in the ellipse, the sum of the squares of any two conjugate diameters is always equal to the sum of the squares of the two axes.

IV°. If we draw ND, the surface of the triangle NCD will be expressed by

(DI+NQ) (CI+CQ)- CI × ID- CQ × NQ = CIxNQ+

CQ×DI=} (NQ*+CQ2) = {x=(a2-CQ2) +CQ2}

Hence the surface of the parallelogram CDEN will be equal to ab, and that of the whole parallelogram FEHG will be equal to 4 ab 2 ax2 b. Consequently all parallelograms circumscribed about an ellipse are equal to each other, and to the rectangle of the two

axes.

35. Call now the semi-diameter CN=m, CD=n, the angle CPM=DCn=P, we shall have I°, m2+n2= a2+b2; 2ly, ab=mn sin P, which is an expression for the surface of the parallelogram CDNE. These two expressions enable us to find immediately the two equal conjugate diameters of an ellipse; for then we have

2 m2 = a2+b2, or m=±√(a2+b2), and sin P=

2 ab a2+b2

And since

these quantities are always real, every ellipse must have two equal conjugate diameters.

With regard to their position, it depends upon the value of CQ; now CQ *+ NQ2, or b2++ CQ2= (a2+b2); therefore CQ=

2

a2+b2

√2,

this value is independant of b, and shews that the ordinate NQ continued, will determine the equal conjugate diameters, in all ellipses, which have the axis Aa in common.

36. Let us now investigate the equation to the co-ordinates CP, PM, and call CP-z, PM=y', NT=q, NQ=r, QT=s, CQ=t,. If we draw PK, MO, perpendicular to the axis, and LP perpendicular to MO; by the similar triangles NQT, MLP, we shall find ML= ry, PL-; and the two other similar triangles CPK, CNQ, give

PK=", CK; hence CO-tz sy, and MO=ry + But

from the property of the ellipse we have MO-a-CO.

Substituting and arranging the terms we have (+y+

(-1)+(+) 2*, and since

az'y' ts) mq

m

ara
12
m2 m2

62

a.

==ts, we shall have (+)*+*+*= a2

(ar
bqq

Observe now that when x'=o, y'=n, therefore the coefficient of

when on the contrary, y'=0, then x'=m, hence the coefficient

n
m2

2

(m2-x*) a result

Consequently the equation becomes y = perfectly agreeing with the equation to the two axes.*

From the above equation it follows that any diameter NCn, bisects the ordinates MPm, and consequently the whole ellipse. And also that every diameter Nn is bisected at the centre C; for at the points Nand n we have z'm2; and therefore 2= ± m.

37. Problem I. Given the two semi-axes a and 6, to find two diameters which shall make a given angle. P=DCn. (See preceding figure.)

ab sin P

We have m2+n2= a2+b2, mn = therefore m2+n2 + 2mn=

2ab sin P

therefore m+n=√(a2+b2+ Consequently m = √(a2+b2+ 2ab sin P

and n

√(a2+b2+

2ab sin P

2ab ; sin P

.) sin P

sin P 2ab ) sin P

5), and m-n= √(a2+b2-200). ) + √(a2+b25) - √(a2+b2-2ab We have now only to determine the direction of one of the diameters, or the angle ACN which call C. The triangle CNT gives m sin P sin (P-C) : m :: sin P: CT= ; hence CQ = CQ sin (P-C) ; therefore in the right angled triangle CNQ we have 1:m::cos C: a* sin (P-C), which gives m2 cos C sin P=a* sin (P

a* sin (P-C) m sin P

aa

m sin P -C) = a2 sin P cos C-a2 sin C cos P, or C cos P, therefore tang C=

a2-m2
a2

tang P.

* Without supposing z' or y'=o, we may equally arrive at the final equation. For Io since a2-t2, the coefficient of za which is

=. Illy. the similar triangles PNQ, DCI give NT* :

m2

CD2 :: QT2: CI2 :: NQ2 : DI2, or q2: n2 :: s2 : a2-t2 :: 12:

38. Problem II. The two diameters m and n being given, and also the angle P which they make with one another; to find the axes and their direction. From the equations mn sin P=ab, and a2+b2 m2+n2, by a similar process to that in the foregoing problems, we deduce a=√(m2+n2+2mn sin P)+√(m2+n2-2mn sin P) and b= √(m2+n2+2mn sin P)-√√(m2+n2-2mn sin P.) The angle C which determines the direction of the two axes, may be found as in the preceding problem.

The figure of which we have now treated, is one of the most useful of all the conic sections. The forms of the orbit in the planatary system are elliptical, and hence a knowledge of its properties is essential in astronomy.

The theory of the ellipse is necessary to establish the projection of the sphere, and the representation of a circle whether othographical or schenographical.

As walls and other objects are frequently built upon elliptical plans, and arches constructed of this figure, to understand the different methods of describing it, under various circumstances, is as necessary to the architect, as it is to the workman employed under him.

It would be impossible to enumerate all the uses to which this beautiful figure can be applied, not only in the research of principles, but in fanciful decoration, which receives much of its elegance from the introduction of the ellipse.