Coroll. Hence AB2: GE2:: CP2-CA2: PM2 (fig. 432.). For let the cutting plane of the opposite hyperbola intersect two circles parallel to the base in HI and PM, and let the cone be cut by another plane parallel to the base, passing through the centre C of the transverse axis, and let mn be the diameter of the circle made by the plane QVN.

Then ACm, APK are similar, and AC: Cm:: AP: PK.
And as BCn, BPL are similar, BC Cn: BP: PL.

Therefore, taking the rectangles of the corresponding terms,
BC AC: Cn x Cm:: BP x AP: PL x PK.

But
BC AC; Cm × Cn= Ct2; and PL x PK = PM2.
Therefore AC2: C:: AP BP: PM2.

Though Ct is not in the same plane, it is what is usually called the semi-conjugate axis, and it agrees with what has been demonstrated in the first part of this proposition.

1086. THEOREM III. In the hyperbola, the square of the semiconjugate axis is to the square of the semi-transverse aris as the sum

A

K

B

R

D

Fig. 432.

N

of the squares of the semi-conjugate aris and of the ordinate parallel to it is to the square of the

abscissas.

Let AB (fig. 433.) be the transverse axis, GE the conjugate, C the centre of the figure, and PM an ordinate, then will

GE2: AB2:: CE2+ PM2: CP2.

For, by Theor. II., CE: CA2:: PM2: CP2 - CA2,
And, by composition, CE2: CA2:: CE2+ PM2 : CP2.

This demonstration may be also applied to what are called conjugate hyperbolas.

1087. THEOREM IV. In the hyperbola, the square of the distance of the focus from the centre is equal to the sum of the squares of the semi-ares.

In

Let AB (fig. 434.) be the transverse axis, CE the semi-conjugate. AB, produced within the curve each way, let F be one focus; and ƒ the other, and let FG be the semi-parameter then CF2CA2+ CE2.

Coroll. 1. The two semi-axes, and the distance of the focus from the centre, are the sides of a right-angled triangle CEA, of which the distance AE

is the distance of the focus from the centre.

Coroll. 2. The conjugate axis CE is a mean proportional between FA and FB, or between ƒB and ƒA, for CE2= CF2-CA=(CF + CA) × (CF— CA)= BF × AF.

1088. THEOREM V. The difference of the radius vectors is equal to the transverse axis.

That is, fM-FM=AB=2CA=2CB.

For

And

CA: CE2:: CP2- CA2: PM2;
CE2-CF2-CA2.

Therefore CA2: CF2-CA2:: CP2-CA2: PM2.

And by taking the rectangle of the extremes and means, and dividing by CA2,

And, subtracting the upper equation from the lower, ƒM— FM=2CA. Coroll. 1. Hence is derived the common method of describing the hyperbolic curve mechanically. Thus : - In the transverse axis AB produced (fig. 435.), take the foci F, f, and any point I in the straight line AB so produced. Then, with the radii AI, BI, and the

centre F, f, describe arcs intersecting each other; call the points of intersection E, then E will be a point in the curve; with the same distances another point on the other side of the axis may be found. In like manner, by taking any other points I, we may find two more points, one on each side of the axis, and thus continue till a sufficient number of points be found to describe the curve by hand. By the same process, we may also describe the opposite hyperbolas.

Coroll. 2. Because

CFX CP
CA

is a fourth proportional, CA, CF, CP CA: CF:: CP: CA+ FM.

1089. THEOREM VI. As the square of the semi-transverse axis is to the square of the semi-conjugate, so is the difference of the squares of any two abscissas to the difference of the squares of their ordinates.

By Theor. II., {

Therefore, by
equality,
And, by division,
Alternately,
But

Therefore

CA: CE:: CP2-CA2: PM2 (fig. 436.),
CA: CE::CH2-CA2: HI2.
CH2-CA2: CP2 - CA2:: HI2: PM2 or
HN2;

CH2-CA2: CH2-CP2:: HI2: HI2-HNo;
CH2-CA2: HI2:: CH⭑- CP2: HIo– HNo.
CH2-CA: HI2:: CA2: CE2

CA2: CE2:: CH2-CP2:: HI2- HN2.

-E

M

P

H N

Fig. 436.

Coroll. 1. If IH be produced to K, and CQ be made equal to CP, then will CH2— CP2=(CH+CP)(CH-CP)=(CP+CH)PH; and HI-HN2=(HI+HN)(HI— HN)=(HI+HN)NI. Therefore the analogy resulting becomes

CA2: CE2::(CP+CH)PH: (HI+HN)NI.

So that the square of the transverse axis is to the square of the conjugate, or the square of the semi-transverse is to the square of the semi-conjugate, as the rectangle of the sum and difference of the two ordinates from the centre is to the rectangle of the sum and difference of these ordinates.

1090. THEOREM VII. If a tangent and an ordinate be drawn from any point in an hyper◄ bola to meet the transverse axis, the semi-transverse axis will be a mean proportional between the distances of the two intersections from the

centre.

CE2: CA2::(IH+HN)IN::(PC+ CH)HP; And by Theor. I., CE2: CA2:: PM2: CP2 — CA2;

For (fig. 437.)

By equality,

PM2 CP2 - CA2:: (IH + HN) IN: (PC +
CH)HP;

And by similar triangles INM, MPT, IN: NM or PH:: PM: PT
or CP-CT.

Therefore, taking the rectangles of the extremes and means of the two last equations, and neglecting the common factors, it will be PM(CP — CT)(CP + CH)=(CP2 — CA2)(IH + HN) ; but when IH and PM coincide, IH and HN each become equal to PM, and CH equal to CP: therefore in the equation substitute 2CP for CP+ CH, and 2PM for IH+HN, and neglecting the common factors and common terms, the result is CT. CP=CA2, or CT: CA::CA: CP.

Fig. 437.

Coroll. Since CT is always a third proportional to CP, CA; suppose the points P and A to remain constant, the point T will also remain constant; therefore all the tangents will meet in the point T which are drawn from the extremity of the ordinate M of every hyperbola described on the same axis AB.

1091. THEOREM VIII. Four perpendiculars to the transverse axis intercepted by it and a tangent, will be proportionals when the first and last have one of their extremities in each vertex, the second in the point of contact, and the third in the centre.

Let the four perpendiculars be AD, PM, CE, BF (fig. 438.), whereof AD and BF have their extremities in the vertices A and B, and the second in the point of contact M of the tangent and the curve, and the third in the centre C.

B

T

D

MA

Fig. 438.

E

But by the similar triangles TAD, TPM, TCE, and TBF, the sides AT, PT, CT, and BT are proportional to the four perpendiculars AD, PM, CE, BF.

Therefore AP: PM::CE: BF.

1092. THEOREM IX. equal angles with a tangent For, by Theor. VII.,

By Cor. 2. Theor. V.,
By equality,

The two radius vectors meeting the curve in the same point will make passing through that point. (Fig. 439.)

CA: CP:: CT: CA;

CA: CP::CF: CA+ FM;
CT: CF:: CA: CA+ FM;

By division and composition, CF-CT: CF+CT:: FM; 2CA

And by the similar triangles TFM, TƒR, FT: ƒT:: FM: ƒR. Therefore ƒR is equal to ƒM; consequently the angle fRM is equal to the angle ƒMR: and because ƒR is parallel to ƒM, the angle FMT is equal to the angle fRM; therefore the angle FMT is equal to the angle ƒRM.

1093. PROBLEM I. To describe an hyperbola by means of the end of a ruler moveable on a pin F (fig 440.) fixed in a plane, with one end of a string fixed to a point E in the same plane, and the other extremity of the string fastened to the other end C of the ruler, the point C of the ruler being moved towards G in that plane.

While the ruler is moving, a point D being made to slide

R

along the edge of the ruler, kept close to the string so as to keep each of the parts CD,

DE of the string taught, the point D will describe

the curve of an hyperbola.

If the end of the ruler at F (fig. 441.) be made moveable about the point E, and the string be fixed in F and to the end C of the ruler, as before, another curve may be described in the same manner, which is called the opposite hyperbola: the points E and F, about which the ruler is made to revolve, are the foci.

There are many occasions in which the use of this conic section occurs in architectural details. For instance, the profiles of many of the Grecian mould. ings are hyperbolic; and in conical roofs the forms are by intersections such that the student should be well acquainted with the methods of describing it.

1094. PROB. II. Given the diameter AB, the abscissa BC, and the double ordinate DE in position and C magnitude, to describe the hyperbola. (Fig. 442.)

Through B draw FG parallel to DE, and draw DF and EG parallel to AB. Divide DF and DC each into the same number of equal parts, Heand from the points of division in BF draw lines to B, also from the points of division in DC draw straight lines to A; then through the points of intersection found by the lines drawn through the corresponding points draw the curve DB. In like manner the curve EB may be drawn so that DBE will form the curve on each side of the diameter AB. If the point A be considered as the vertex, the opposite hyperbola HAI may be described in the same manner, and thus the two curves formed by cutting the opposite cones by the same plane will be found. By the theorists, the hyperbola has been considered a proper figure of equilibrium for an arch whose office is to support a load which is greatest at the middle of the arch, and diminishes towards the abutments. This, however, is matter of consideration for another part of this work.

F.

3

2

D1 2 3 C

Fig. 442.

E

OF THE PARABOLA.

1095. DEFINITIONS.-1. The parameter of the axis of a parabola is a third proportional to the abscissa and its ordinate.

2. The focus is that point in the axis where the ordinate is equal to the semi-parameter. 3. The diameter is a line within the curve terminated thereby, and is parallel to the axis.

4. An ordinate to any diameter is a line contained by the curve and that diameter parallel to a tangent at the extremity of the diameter.

1096. THEOREM I. In the parabola, the abscissas are proportional to the squares of their

ordinates.

K

V

Let QVN (fig. 443.) be a section of the cone passing along the axis, and let the directrix RX pass through the point Q perpendicular to QN, and let the parabolic section be ADI meeting the base QIND of the cone in the line DI, and the diameter QN in the point H; also let KML be a section of the cone parallel to the base QIN intersecting the plane VQN in the line KL, and the section ADI in PM. Let P be the point of concourse of the three planes QVN, KML, AHI, and let H be the point of concourse of the three planes QVN, KML, AHI; then, because the planes VRX and ADI are parallel, and the plane AQN is perpendicular to the plane ARX, the plane ADI is also perpendicular to the plane AQN. Again, because the plane QIN is perpendicular to the plane QVI, and the plane KML is parallel to the plane QIN, the plane KML is perpendicular to the plane QVN: therefore the common sections PM and III are perpendicular to the plane AQN; and because the plane KML is parallel to the plane QIN; and these two planes are intersected by the plane QVN, their common sections KL and QN are parallel. Also, since PM and HI are each perpendicular to the plane QVN, and since KL is the common section of the planes QVN, KML, and QN in the common section of the planes QVN, QIN; therefore PM and HI are perpendicular respectively to KI and QN.

Consequently

AP: AH:: PM2: HI2.

For, by the similar triangles APL, AHN, AP: AH:: PL: HN,

Or

But, by the circle

And, by the circle

Therefore

Therefore, by substitution,

R

Q

M

Coroll. By the definition of the parameter, which we shall call P,

1097. THEOREM II. As the parameter of the axis is to the sum of any two ordinates, so is the difference of these ordinates to the difference of their abscissas.

That is, P: HI+ PM :: HI-PM; AH-AP.

Multiplying the first of these equations by AP and the second by AH,

Subtract the corresponding numbers of the first equation, and P (AH-AP) = HI2- PM2. But the difference of two squares is equal to a rectangle under the sum and difference of their sides.

And

HI2- PM2 (HI + PM) (HI – PM).

Therefore P(AH-AP) = (HI + PM) (HI-PM).
Consequently P ; HI+ PM :: HI-PM: AH-AP;

Or, by drawing KM parallel to AH, we have GK = PM + HI, and KI=HI-PM; and since PH AH-AP; P: GK::KI: PH, or KM.

Therefore, by multiplication, KM × HI2= GK × KI× AH, or
AH: KM::HI2: GK × KI.

So that any diameter MK is as the rectangle of the segments GK,
KI of the double ordinate GI. From this a simple method has been
used of finding points in the curve, so as to describe it.

1098. THEOREM III. The distance between the vertex of the curve and

the focus is equal to one fourth of the parameter.

Let LG (fig. 445.) be a double ordinate passing through the focus, then LG is the parameter. For by the definition of parameter AF: FG:: FG: P=2FG.

Therefore 2AF=FG={LG;

Consequently AF = {LG.

1099. THEOREM IV. The radius vector is equal to the sum of the distances between the focus and the vertex, and between the ordinate and the vertex. (Fig. 446.)

Therefore

FP2

But, by Cor. Theor. II.,
Therefore, by addition,

H

AP2-2AP × AF + AF2.
PMP x AP=4AF x AP.
FP2+ PM2 AP2 + 2AF × AP
+ AF2.

But by the right-angled tringles, FP2 + PM2 = FM2;
And therefore

Hence, extracting the roots,

Or by making

FM2 AP2 +2AFx AP + AF2.

-

FM AP+AF=2AF+ FP;
AG=AF, FM=GP.

Coroll. 1. If through the point G (fig. 447.) the line GQ be drawn perpendicular to the axis, it is called the directrix of the parabola.

By the property shown in this theorem, it appears that if any line QM be drawn parallel to the axis, and if FM be joined, the straight line FM is equal to QM; for QM is equal to GP.

T

Coroll. 2. Hence, also, the curve is easily described by points. Take AG equal to AF, (fig. 447.), and draw a number of lines M, M perpendicular to the axis AP; then with the distances GP, GP, &c. as radii, and from F as a centre, describe arcs on each side of AP, cutting the lines MM, MM, &c. at MM, &c.; then through all the points M, M, M, &c. draw a curve, which will be a parabola.

1100. THEOREM V. If a tangent be drawn from the vertex of an ordinate to meet the axis MK produced, the subtangent PT (fig. 448.) will be equal to twice the distance of the ordinate

from the vertex.

If MT be a tangent at M, the extremity of the ordinate PM; then the sub-tangent PT is equal to twice AP. For draw MK parallel to

But when the ordinates HI and PM coincide, MT will become a tangent, and GK will become equal to twice PM.

From this property is obtained an easy and accurate method of drawing a tangent to any point of the curve of a parabola. Thus, let it be required to draw a tangent to any point M in the curve. Produce PA to T (fig. 449.), and draw MP perpendicular to PT, meeting AP in the point P. Make AP equal to AP, and join MT, which will be the tangent required.

1101. THEOREM VI. The radius vector is equal to the distance between the focus and the intersection of a tangent at the vertex of an ordinate and the axis produced. Produce PA to T (fig. 450.), and let MT be a tangent at M; then will FT FM.

Coroll. 1. If MN be drawn perpendicular to MT to meet the axis in N, then will FN=FM=FT. For draw FH perpendicular to MT, and it also bisects MT, because FM= FT; and since HF and MN are parallel, and MT is bisected in H, the line TN will also be bisected in F. It therefore follows that FN=FM=FT.

Coroll. 2. The subnormal PN is a constant quantity, and it is equal to half the parameter, or to 2AF. For since TMN is a right angle,

Therefore

2AP or TP: PM:: PM: PN.

But, by the definition of parameter, AP: PM:: PM; P;
Therefore

PN={P.