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CF.CP

Extracting the root from each number, FM CA- СА

In the same manner it may be shown that FM=CA+ these is FM+M=2CA.

CF.CP
; therefore the sum of
CA2

Coroll. 1. A line drawn from a focus to a point in the curve is called a radius vector, and the difference between either radius vector and half the major axis is equal to half the difference between the radius vectors. For, since

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Coroll. 3. Hence the difference between the major axis and one of the radius vectors gives the other radius vector. For, since FM + Mƒ=2CA;

Therefore FM=2CA-Mƒ.

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h

Coroll. 4. Hence is derived the common method of describing an ellipsis mechanically, by a thread or by points, thus: - Find the foci Ff (fig. 414.), and in the axis AB assume any point G; then with the radius AG from the point F as a centre describe two arcs H, H, one on each side of the axis; and with the same radius from the point ƒ describe two other arcs h, h, one on each side of the major axis Again, with the distance

A

G

H

h

Fig. 414.

B

GB from the point ƒ describe two arcs, one on each side of the axis, intersecting the arcs HH in the points HH; and with the same radius from the point ƒ describe two other arcs, one on each side of the axis, intersecting the arcs described at h, h in the point h, h. find as many points as we please; and a sufficient number being found, the curve will be formed by tracing it through all the points so determined.

In this manner we may

1070. THEOREM VI. The square of half the major axis is to the square of half the minor axis as the difference of the squares of the distances of any two ordinates from the centre to the difference of the squares of the ordinates themselves.

Let PM and HI (fig. 415.) be ordinates to the major axis AB; draw MN parallel to AB, meeting HI in the point N; then will PM=HN, and MN=PH, and the property to be demonstrated is thus expressed

CA: CE:: CP-CH: HJ2- HNo.

I E

M

N

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Or by producing HI to meet the curve in the point K, and making CQ = CP, the property to be proved will be

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Therefore

But, by division,

Alternately,

CA2-CH2: CA2-CP2::HI2: PM2 or HN2;
CA2-CH2: CP2-CH2:: H12: HI-HNo.
CA2-CH2: HI2:: CP2 – CH2; HI2 – HN2;

And, since we have above, CA- CH2: HI2:: CA: CE,
Therefore, by equality,

But since

And since
Therefore

CA: CE2:: CP2- CH2; HI2- HN2;
CP_CH2=(CP-CH)CP+CH)=PH × QH,
HIo — I N2 = (HI − HN)(HI + HN)= NI × KN,
CA: CE:: PH x HQ: NI NK.

Coroll. 1. Hence half the major axis is to half the minor axis, or the major axis is to the minor axis, as the difference of the squares of any two ordinates from the centre is to the rectangle of the two parts of the double ordinate, which is the greatest made of the sum and difference of the two semiordinates. For KN=HK + HN= HI+HN, which is the sum of the two ordinates, and NI-HI-HN, which is the difference of the two ordinates. Coroll. 2. Hence, because CP2-CH2=(CP-CH)(CP+ CH), and since HI— HN? = (HI-HN)(HI+HN), and because CP-CH= PH and HI-HN=NI; therefore CA:CE::(CP+CH)PH ;(HI+HIN)NI

X

1071. THEOREM VII. In the ellipsis, half the major axis is a mean proportional between the distance of the centre and an ordinate, and the distance between

the centre and the intersection of a tangent to the vertex of that ordinate.

To the major axis draw the ordinates PM (fig. 416.) and HI, and the minor axis CE. Draw MN perpendicular to HI. Through the two points I,M. draw MT, IT, meeting the major axis produced in T; then will CT; CA::CA: CP. For,

E

N

B

4 P

H

C

Fig. 416.

By Cor. 1. Theor. VI., CE2: CA2::(IH + HN)IN: (PC+ CH)HP;
By Cor. 2. Th. II., CE2: CA2:: PM2: CA-CP2;

Therefore, by equality, PM2: CA2 – CP2 : : (IH +HN)IN : (PC + CH)HP. By similar triangles, INM, MPT; IN: NM or PH:: PM: PT or CT-CP. Therefore, taking the rectangles of the extremes and means of the two last equations, and throwing out the common factors, they will be converted to the equation

PM(CT-CP)(CP+CH)=(CA2 — CP2)(IH + HN).

But when HI and PM coincide, HI and HN will become equal to PM, and CII will become equal to CP; therefore, substituting in the equation 2CP for CP+ PH, and 2PM for IH+HN, and throwing out the common factors and the common terms, we have

CT. CP CA2

or CT: CA:: CA: CP.

Coroll. 1. Since CT is always a third proportional to CP and CA, if the points P, A, B remain fixed, the point T will be the same; and therefore the tangents which are drawn from the point M, which is the intersection of PQ and the curve, will meet in the point T in every ellipsis described on the same axis AB.

Coroll. 2. When the outer ellipsis AQB, by enlarging, becomes a circle, draw QT perpendicular to CQ, and joining TM, then TM will be a tangent to the ellipsis at M.

Coroll. 3. Hence, if it were required to draw a tangent from a given point Tin the prolongation of the major axis to the ellipsis AEB, it will be found thus: - On AB describe the semicircle AQB. Draw a tangent TQ to the circle, and draw the ordinate PQ intersecting the curve AEB of the ellipsis in the point M; join TM; then TM is the tangent required. This method of drawing a tangent is extremely useful in practice.

1072. THEOREM VIII. Four perpendiculars to the major axis intercepted by it and a tangent will be proportionals when the first and last have one of their extremities in the vertices, the second in the point of contact, and the third in the centre.

Let the four perpendiculars be AD, PM, CE, BF, of which AD and BF have their extremities in the vertices A and B, the second in the point of contact M, and the third in the centre C; T then will

AD: PM::CE: BF.

For, by Theor. VII., TC: AC:: AC: CP;

F

E

M

D

P

By division,

That is,

By composition,

Therefore

Fig. 417.

TC-AC CA-CP::TC: AC or CB;
TA: AP::TC: CB.

TA: TA+AP::TC: TC + CB:

TA: TP::TC: TB.

But by the similar triangles TAD, TPM, TCE, and TBF, the sides TA, TP, TC, and TB are proportionals to the four perpendiculars AD, PM, CE, and BF; therefore

AD: PM::CE: BF.

Coroll. 1. If AM and CF be joined, the triangles TAM and TCF will be similar. For by similar triangles, the sides TD, TM, TE, TF are in the same proportion as the sides TA, TP, TC, TB.

Therefore TD: TM::TE: TF;

Alternately, TD: TE::TM. TF: but TAD is similar to TCE;
Hence

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Therefore, by equality, AP: PM:: CB: BF.

Coroll. 3. If AF be drawn cutting PM in I, then will PI be equal to the half of PM

For, since AP: PM:: CB: BF, and, by the similar triangles API, ABF,
AP: PI::AB: BF;

Therefore PM: PI::CB; AB.

But CB is the half of AB; therefore, also, PI is the half of PM.

1075, THEOREM IX. If two lines be drawn from the foci of an ellipse to any point in the curve, these two lines will make equal angles with a tangent passing through that point.

Let TM (fig. 418.) be a tangent touching the curve

at the point M, and let F, ƒ be the two foci; join
FM, ƒM. then will the angle FMT be equal to the
angle ƒMR. For draw the ordinate PM, and draw
fR parallel to FM, then will the triangles TFM and T
TfR be similar; and by Cor. Theor. VII.,

By Cor. 2. Theor. V.,

Therefore, by equality,

By division and composition,
That is,

CA CP:: CT: CA;

G

D

M

CA: CP::CF: CA-FM;
CT: CF:: CA: CA-FM.

AF PNC

Fig. 418.

CT-CF: CT+CF:: FM: 2CA- FM;
TF: Tƒ:: FM : ƒM.

By the similar triangles TFM, TƒR; TF: Tƒ:: FM : ƒR.

It therefore appears that ƒM is equal to fR, therefore the angle fMR is equal to the angle fRM: but because FM and ƒR are parallel lines, the angle FMT is equal to the angle fRM; therefore he angle FMT is equal to the angle ƒ MR.

Coroll. 1. Hence a line drawn perpendicular to a tangent through the point of contact will bisect the angle FMf, or the opposite angle DMG. For let MN be perpendicular to the tangent TR. Then, because the angle NMT and NMR are right angles, they are equal to one another; and since the angles FMT and ƒMR are also equal to one another, the remaining angles NMF and NMƒ are equal to one another. Again, because the opposite angles FMN and IMG are equal to one another, and the opposite angle ƒMN and IMD are equal to one another; therefore the straight line MI, which is the line MN pro. duced, will also bisect the angle DMG.

Coroll. 2. The tangent will bisect the angle formed by one of the radius vectors, and the prolongation of the other. For prolong FM to G. Then, because the angles RMN and RMI are right angles, they are equal to one another; and because the angles NMƒ and IMD are equal to one another, the remaining angles RMG and RMƒ are equal to one another.

Scholium. Hence we have an easy method of drawing a tangent to any given point M in the curve, or of drawing a perpendicular through a given point in the curve, which is the usual mode of drawing the joints for masonic arches. Thus, in order to draw the line IM perpendicular to the curve: produce FM to G, and ƒM to D, and draw MI bisecting the angle DMG; then IM will be perpendicular to the tangent TR, and consequently to the

curve.

As in optics the angle of incidence is always found equal to the angle of reflection, it appears that the foundation of that law follows from this theorem; for rays of light issuing from one focus, and meeting the curve in any point, will be reflected into lines drawn from these points to the other focus: thus the ray ƒM is reflected into MF: and this is the reason why the points Fƒ are called foci, or burning points. In like manner, a sound in one focus is reflected in the other focus.

K

E

M

A H

P

B

1074. THEOREM X. Every parallelogram which has its sides parallel to two conjugate diameters and circumscribes an ellipsis is equal to the rectangle of the two axes. Let CM and CI (fig. 419.) be two semi conjugate diameters. Complete the parallelogram CIDM. Produce CA and MD to meet in T, and let AT meet DI in t. Draw IH and PM ordinates to the axis, and draw half the minor T axis CE. Produce DM to K, and draw CK perpendicular to DK then will the parallelogram CIDM be equal to the rectangle, whose sides are CA and CE; or four times the rectangle CIDM will be equal to the rectangle made of the two axes AB and GE

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Fig. 419.

G

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But by the similar triangles HIC, KCT, HI: CI::CK: CT;

Therefore

Consequently

CE: HI:: CT: CA.

CE: CI:: CK: CA:

CEx CACI x CK.

The ellipsis is of so frequent occurrence in architectural works, that an acquaintance with all the properties of the curve, and the modes of describing it, is of great importance to the architect. Excepting the circle, which may be called an ellipsis in which the two foci coincide, it is the most generally employed curve in architecture.

1075. PROBLEM I. To describe an ellipsis.

F

D

G

Let two pins at E and F (fig. 420.) be fixed in a plane within a string whose ends are made fast at C. If the point C be drawn equally tight while it is moved forward in the plane till it returns to the place from which it commenced, it will describe an ellipsis.

1076. PROB. II. The two diameters AB and ED of an ellipse being given in position and magnitude, to describe the curve through points.

Fig. 120.

E

Fig. 421.

Let the two diameters cut each other at C (fig. 421.). Draw AF and BG parallel to ED. Divide AC and AF each into the same number of equal parts, and draw lines, as in the figure, through the points of division; viz. those from the line AF to the point D, and the lines through AC to the point E; then through the points of intersection of the corresponding lines draw the curve AD, and in the same manner find the curve BD; then ADB will be the semi-ellipsis.

D

It is evident that the same method also extends to a circle by making CD equal to CA; (fig. 422.); and it appears that the two lines forming any point of the curve to be drawn will make a right angle with each other. For these lines terminate at the extremities of the diameter ED, and the point of concourse being in the curve, the angle made by them must be a right angle; that is, the angle EAD, or EhD, or E¿D. or EkD. is a right angle: and from this property we have the following method of drawing the segment of a circle through points found in the curve.

23

Fig. 122.

Thus, let AB be the chord, and CD be the versed sine of an arc of a circle, to describe the arc. Through D draw H1 (fig. 423.) parallel to AB; join AD and DB; draw AH perpendicular to AD, and BI perpendicular to BD; divide AC and HD each into the same number of equal parts, and join the corresponding points; divide AF into the same number of equal parts, and through the points of division draw lines to D, and through the corresponding points where these lines meet the former draw a curve AD. In the same manner the other half BD may be drawn.

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1077. PRCS. III. A diameter KH of an ellipsis being given, and an ordinate DL, to find the limits of the other conjugate diameter.

Bisect KH in (fig. 424.), through I draw EA parallel to DL, and draw DC and KB perpendicular to EA; from the point L with the distance K describe an arc cutting EA at F; join LF, and produce LF to C; make IE and IA each equal to LC; then will EA be a diameter conjugate to KH.

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1078. PROB. IV. A diameter KH and an or linate DL of an ellipsis being given, to describe the curve. (fig. 424.)

H

Fig. 421.

Find the limits E and A of the other conjugate diameter by the preceding construction. Produce KB to q, and make Kq equal to 1A or IE, and through the centre 1 of the curve and the point q, draw the straight line MN. Then, suppose the straight line KBq to be an inflexible rod, having the point B marked upon it. Move the rod round, so that the point q on the rod may be in the line MN, while the point B is in the line EA: then, at any instant of the motion, the place

of the point K on the plane whereon the figure is to be drawn may be marked; the points thus found will be in the curve. Instead of a rod, a slip of paper may be used, and in some cases a rod with adjustible points to slide in a cross groove, and a sliding head for a pencil is convenient; and such an instrument is called a trammel.

K

L

DI

F

When the diameters KH and EA (fig. 425.) are at right angles to each other, the straight line Kq coincides with the diameter KH, and consequently the line MN, on which the point q of the inflexible line Kq moves, will also fall upon the diameter KH. Therefore in this case nothing more is required to find the limits of the other diameter, than to take the half diameters IK, KH of the given diameters, and from the extremity L with that distance describe an arc cutting the unlimited diameter in the point F; then drawing LF, and producing it to q, and making IE and IA each equal to qL, EA will be the other diameter; and since the two diameters are at right angles to each other, they are the two axes given in position and magnitude, and thus the curve may be described as before.

Fig. 425.

A method of describing the curve from any two conjugate diameters is occasionally of considerable use, and particularly so in perspective. For, in every representation of a circle in perspective, a diameter and a double ordinate may be determined by making one of the diameters of the original circle perpendicular to the plane of the picture and the other parallel to it; and then the representation of the diameter of the original circle, which is perpendicular to the intersecting line, will be a diameter of the ellipsis, which is the representation of that circle; and the representation of the diameter of the circle which is parallel to the intersecting line will become a double ordinate to the diameter of the ellipsis which is the perspective representation of the circle.

B

1079. PROB. V. Through two given points A and B to describe an ellipsis, the centre C being given in position and the greater axis being given in magnitude only. About the centre C (fig. 426.) with a radius equal to half the greater axis describe a circle HEDG; join AC and BC; draw AD perpendicular to AC, and BE perpendicular to BC, cutting the circumference in the points D and E; draw also BF parallel to AC, and find BF, which is a fourth proportional to AD, AC, and BE; through the point F and the centre C draw FG to cut the circle in H and G, and GH is the major axis of the ellipsis. By drawing an ordinate Bq, the curve may be described by the preceding problem, having the axis GH and the ordinate Bq.

C

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H

E

Fig. 426.

1080. PROB. VI. Through a given point in the major axis of a given ellipsis to describe nother similar ellipsis which shall have the same centre and its major axis on the same straight line as that of the given ellipsis.

A

C

K

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B

Let ACBD (fig. 427.) be the given ellipsis, having AB for its major axis and CD for its minor axis, which are both given in position and magnitude. It is required to draw a similar ellipsis through the point G in the major axis AG. Draw BK perpendicular and CK parallel to A B, and join KE. Again, draw GL perpendicular to AB cutting EK at L, and draw LH parallel to AB cutting CD in H. On the axis CD make EI equal to EH, and on the axis AB make EF equal to EG. Then, having the major axis AB, and the minor axis FG, the ellipsis FIGH may be described, and when drawn, it will be similar to the given ellipsis ADBC.

།།

D

Fig. 127.

1081. PROB. VII. Through any given print p, within the curve of a given ellipsis to describe another ellipsis which shall be similar and concentric to the given one.

Let C (fig. 428.) be its centre. Draw the straight line CpP, cutting the curve of the

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given ellipsis in P. In such curve take any other number of
points Q, R, S, &c., and join QC, RC, SC, &c.; join PQ and
draw pq parallel thereto cutting qC at q: join PR and draw pr
parallel to PR, cutting RC at r; join PS and draw ps parallel to r
PS cutting SC in s. The whole being completed, and the curve
p, s, t, u drawn through the points p, q, r, s, &c., the figure will
be similar and concentric to the given ellipse P, S, T, U; or when
the points at the extremities for one half of the curve have been
drawn, the other half may be found by producing the diameter to the opposite side, and
making the part produced equal to the other part.

S

Fig. 428

1082. PROB. VIII. About a given rectangle ABCD to describe an ellipsis which shall have its major and minor axes respectively parallel to the sides of the rectangle and its centre in the points of intersection of the two diagonals.

Bisect the sides AD and AB (fig. 429.) of the rectangle respectively at L and O

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