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2 ay

aq

2 apy ✓ aq

To find the equation of the co-ordinates of the diameter MO, call MP, &'; PN, y; AQ=AT=a; we shall have MQ=vap, q=p.+4a, MT=V ag; and if we draw NL perpendicular to the axis, the similar triangles NRL, MTQ, will give MT: MR :: MQ: NL

::QT:RL

2 ay or / aq=y+w ag:: vap:NL-' Vap

tap::2 a:RL= +2a saq

Vaq But AR-RT-AT-r'-; therefore AL=x'+a+ and by the property of the parabola, NL’=PX AL, or (vap + y vap)'=ap+px' + .; from which by reduction we

saq obtain y'y=qx', an equation similar to that which was before found for the co-ordinates of the axis. Hence we may conclude that any diameter MO bisects all the ordinates Nn.

The two following problems afford an easy application of the above principles.

25. I. Given the axis AL of a parabola and its parameter p, to find a diameter MO, which shall make with its ordinates a given angle MPn=-A. (See preceding figure).

This problem merely requires us to find the point Q where the perpendicular MQ meets the axis. For this purpose call AQ-A, the triangle MTQ will give

AT : MQ:: Radius : tang A (because z MPn=4MTQ) or in symbols, 2 ri vpr::1 : tang A Hence x =

P

4 tang? A=4 p cot ’A, and the parameter of the diameter MO, or q=p+4 x=P(1 + cot? A)=p cosec ?A=_P.

It is easy to see that this problem has two solutions.

Prob. II. Given the parameter q of the diameter MO, with the vertex M of that diameter, and the angle A which it makes with its ordinates; to find the axis AL, its vertex A, and its parameter p. (See preceding figure.)

This problem requires us to find the distance MQ from the axis to the diameter, and then the distance AQ, in order to obtain the vertex A and the parameter p. Preserving the same denominations as in the preceding problem, we have MQ=vpe,

р q=p+ 42=

Hence we obtain p==q sin ’A, x=\ q cos ’A,

sin? A and MQ=+} q sin A cos A=+* q sin 2 A.

The properties of the parabola are of frequent and great use in the Arts and Sciences.

sin ?A

OF THE ELLIPSE.

cx

sin B—xx sin (A+B)}

BY

M

The equation of the Ellipse is

sin A yy=

cos? | BI Consequently to every abscissa

Ő AP correspond two ordinates

N

B PM, PM' equal and opposite to each other. If we make y=0, we shall obtain the two points in which the curve meets the axis of the abscissæ, that is the trans А verse axis Aa. The first of these points is at A, where x=0; the

Z

csin B second ata, where x=.

sin (A + B) as we shall find by resolving the equation cx sin (A+B=0. The constant quantity

I suppose equal

sin (A+B) to the transverse axis Aa. Call this transverse axis 2 a, and we shall have

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P

IM Z

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sin

B - XX

c sin B

sin A sin (A+B) (2 ax—xx)

yy

cos? 1 B 26. The double ordinate BC6 passing through the middle C of the axis Aa, or through the center of the ellipse is called the less or conjugate axis. To introduce it into the equation of the ellipse, call it 2 b; and since when y=b, x=a, we shall have sin A sin (A+B)

Hence by substitution we find cosB bb yy=-(2 ar-xx)

bb

aa.

aa

This expression gives the following proportion yy: 2 ax-xx::66 : aor PM? : AP x Pa::CB?: CA’; that is, in the ellipse the squares of the ordinates of the transverse aris are to the products of their corresponding abscissæ, as the square of the conjugate or lesser axis is to the square of the transverse axis.

aa

If we describe a circle whose center is C and radius CA, we shall have PN=APx Pa. Therefore PN : PM::a: 6::CB' : CB

Hence the ordinates of the ellipse are proportional to the ordinates of a circle described on the transverse axis ; this gives us an easy method of describing an ellipse. It is only necessary for this pure pose to describe a curve through a series of points taken on the ordinates of a circle cut into similar parts.

27. If we had reckoned the absciscæ from the center C, making CP=2, we have r=1-2, and substituting this value in the equation above found, we have bb

62 22 yy=-(aa-z) = -; an equation to which we shall often

al refer.

If b were equal to a, we should have yy=aamzz an equation of the circle ; hence we may consider the circle as an ellipse whose two axes are equal.

bb The equation yy=% (aa-22) gives

zz: bb-yy::aa : bb, or MQ’: BQ x Qb::CA?: CB?. Therefore in the ellipse the squares of the ordinates of the lesser axis are to the rectangles of their abscissæ as the square of the transverse is to the square of the conjugate axis.

28. If from one of the extremities B of the conjugate axis,

N and with a radius BF equal to

B the semi-transverse axis CA, we describe a circular arc, it will intersect the transverse axis in two points F and f, which are А) called the Foci. From this it appears that the distance CF is equal to (aa-bb), hence

M 13 AF X Fa=

aa

M м

P

Fa={a-vlaa-bb)} {a+

+v(aa-bb)}=bb=CB*.

Hence the semi-conjugate axis is a mean proportional between the distances from either of the foci to the two vertices of the ellipse.

29. The ordinate DF passing through the focus has for its expression; and Dd, which is called the parameter p of the trans

2 lb 4 b6 verse aris

Consequently 2 a : 26 :: 26:p, and

2 u therefore the parameler is a third proportional to the transverse and conjugate axis. By analogy to this properly we call the parameter

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=

a

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we shall have fin+mF=FM+Mf=or 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* =mM=LMT. Therefore if we prolong the radius vector f M, the line MT which bisects the angle LMT, will be the tangent required.

The angle LMT=OM=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+ LEMN=Z TMF+ZNMF, and we have just shewn that ZOMF, or its equal LMT=TMF) therefore we shall have fM:FM::fN: FN (Euclid), or by composition

fM+FM : FM ;: N+FN : FN or,

2 a

2 C

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C

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or since (?=a2—62, by substitution we have FN=C-%+

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Consequently FN + --C=PN=

pp. This is the expression

2 a 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

рх sa—%, we should have PN=

3 p

2 a 64 22

62 22 The normal NM=(yy +

62 (a at

bb

bb x

a

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2 ax-XX (When the origin of the abscissæ is at the vertex PT

) Consequently CT=PT+CP=.

P="", which gives this proportion

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 ImT, FMT may be considered as equal.

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