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For, let XI be drawn and produced to meet the other assymptote in Z. Then, because XL is equal and parallel to ÍC, XZ is parallel to LM. Now, the triangles ICZ, LCY, by reason of parallel lines, are equiangular, and they have the side IC = LY; therefore IZ = LC = IX. Whence XZ is a tangent at I; and LM has beeen proved parallel to it. Consequently, IN, LM, are mutually parallel to each other's ordinates. Cor: 1.-Tangents to
the four hyperbolas, at the vertices of two conjugate diameters, form a parallelogram, whose diagonals are
coincident with the assymptotes. CoR. 2.-Any two conjugate diameters of an equilateral
hyperbola are equal to one another. For XCY is in this case a right-angled triangle, and therefore L is the centre of its circumscribing circle, and CL = LX = CI, or ML = IN.
PROPOSITION XII. If an ordinate be applied to any diameter, the square of that diameter is to the square of its conjugate as the rectangle under its segments to the square of the semi-ordinate.
Let LCM be any diameter, IN its conjugate, and DRO an ordinate to LM, in the hyperbola DLO; LMP: IN = MR·RL: DRP (figure to proposition 11).
For, let DO meet the assymptotes in Q, V, and the tangent at L meet them in X, Y. Then, from similar triangles, CL : CR2 = LX? : RQ%, by conversion and alternation, CL2 : LX? = MR·RL: RQ2 · LX2. But LX2 = CI? (III. 9, Cor.3)=QDDV (III. 9), = RQ— DR?, hence ŘQ LX=DR2; therefore CL: CI? = MR. RL: DR, and LM2: IN= MRRL:DR.
Cor. 1.-If an external ordinate be applied to any dia
meter, the square of that diameter is to the square of its conjugate, as the sum of the squares of the semidiameter and segment intercepted by the external ordinate, from the centre, to the square of the external
semi-ordinate. For, let TW be an external ordinate; then CL:CU2 = LX2: UP2 ; by addition CL: CL? + CU2 = LX2: LX? + UP2; by alternation CL : LX2 = CL + CU? : LX? + UP2. Now LX2 = CI2 = PT: TS=UT? — UP2; therefore UTP = LX? + UP, and hence CL : CI2 = CL + CUP: UT.
Schol. The proof in this corollary is the same as that of the proposition, with the exception that addition is taken instead of conversion. CoR. 2.—The squares of semi-ordinates are to one an
other as the rectangles under the segments of the
diameter. CoR. 3.-Any diameter of an hyperbola is to its para
meter, as the rectangle under its segments to the square
of the semi-ordinate that divides them. For, LM:IN=IN :P, the parameter; then LM:P= LM: IN2 = CL: CI2 = MR. RL: DR. Cor. 4.-In an equilateral hyperbola, the rectangle under
the segments of any diameter is equal to the square of the semi-ordinate that divides them; and the square of an external semi-ordinate is equal to the sum of the squares of the semi-diameter and segment intercepted
from the centre. For then CL = CI, and therefore MR·RL=DR2, and CL? + CUP=UT.
PROPOSITION XIII. If from any point in an hyperbola, two straight lines be drawn, to meet the assymptotes, and from any other point in the same, or in the opposite hyperbola, straight lines, parallel to these, be also drawn to the assymptotes; the rectangle under the former shall be equal to the rectangle under the latter.
From the point D, in the hyperbola, let there be drawn
any two straight lines DQ, DR, to terminate in the assymptotes CQ, CS, and from any other point N let NP be drawn parallel to DQ, and NS parallel to DR, to terminate in the same lines. Then RD:DQ = PN.NS.
For, let LDK and MNO be drawn parallel to the conjugate axis, to meet the assymptotes. Then, from the similar triangles RDK, SNO, RD: DK = NS: NO, and from the similar triangles, LDQ, MNP, DQ : LD = PN : MN. Therefore RD · DQ : LD.DK = PN: NS: MN: NO. But LD.DK = MN • NO (III. 9, Cor. 4). Consequently RD · DQ = PN.NS. Cor. 1.—If from two points in the same or in opposite
hyperbolas, straight lines be
tercept from the centre. For QĪ, DR, being parallel respectively to PN and NS, therefore QD · DR NS · NP, and consequently QD : NS = NP: DR, or QD: NS = CS: CQ. Cor. 2.-If from any point in a given hyperbola, two
straight lines be drawn parallel to the assymptotes, the
parallelogram formed thereby is of a given magnitude. For (Pl. Ge. VI. 23, Cor. 1) QR : PS = (DQ : NS, DR:NP)=DQ · DR:NSNP; but DQ-DR=NS.NP, therefore QR =PS, and any other rectangle similarly formed is shown in the same manner to be equal to PS. COR. 3.-Every sector of an hyperbola is equal to the
quadrilateral figure contained by the curve, by one assymptote, and by parallels to the other, through the
extremities of the base of the sector. For, since the parallelograms RQ, PS, are equal, the two
triangles CQD, CSN, are together equal to PS, and these equals being taken from the figure CQDNS, there remains the sector ČDN equal to the quadrilateral figure PQDN.
PROPOSITION XIV. If in one of the assymptotes of an hyperbola, any number of points be assumed, such, that their distances from the centre be in continued proportion, and straight lines be drawn from these points to the curve, parallel to the other assymptote, the mixtilineal quadrilateral figures formed thereby will be equal.
Let the points A, B, D, E, be assumed in the assymptote CS, so that CA: CB = CB:CD = CD: CE, and let AF, BG, DH, EK, be drawn to meet the curve parallel to the other as- Q symptote CQ; the quadrilateral figures AFGB, BGHD, DHKE, shall be equal.
For, let the tangent at G, and the line that joins H, F, meet CQ in N, L, C and CS in 0, M. Then, because OG = GN (III. 9, Cor. 3), OB = BC, and because MH : FL (III. 9, Cor. 2), MD = AC; therefore MD:OB=CB: CD = DH: BG (III. 13, Cor. 1). Hence, the triangles MDH, OBG, which have the angles at D and B equal, are equiangular, and LM is parallel to NO. The diameter CG, therefore, bisects the chord HF, and every chord parallel to HF (III. 10). Consequently CG bisects the segment FGH of the hyperbola. But it also bisects the triangle FCH; therefore the sector CFG=CGH, and the quadrilateral AFGB = BGHD (III. 13, Cor. 3).
Schol. ---The ninth and twelfth propositions of the ellipse may be applied to the hyperbola, and demonstrated in the
EXERCISES. 1. The square of any semi-diameter of an hyperbola is equal to the rectangle under the distances of its vertex from the foci, added to the difference of the squares of the semitransverse and semi-conjugate axis.
2. Every tangent of an hyperbola is harmonically divided by the transverse axis and perpendiculars falling upon it from the foci.
3. The difference of the squares of any two conjugate diameters of an hyperbola, is equal to the difference of the squares of the two axes.
4. If from any point in an hyperbola, straight lines be drawn through the vertices of a diameter, to limit the tangents at these points, the rectangle under the tangents will be equal to the square of the semi-conjugate diameter.
5. A semi-ordinate to any diameter is a mean proportional between its segments, intercepted from the diameter by two straight lines intersecting each other in any point of the curve, and passing through the vertices of the diameter.
6. If a quadrilateral figure be formed by tangents to the four hyperbolas, a straight line through the centre, parallel to that which joins two opposite points of contact, will divide the two opposite sides of the figure, so that the segments of the one shall be inversely proportional to the segments of the other.
7. Also, the straight line which joins the middle points of its diagonals will pass through the centre of the hyperbolas.
8. If through a fixed point, any straight line be drawn, to meet the hyperbola, or opposite hyperbolas, in two points, the rectangle under its segments, from the fixed point, will be to the rectangle under its segments, intercepted by the assymptotes, from either point in the curve, in a constant ratio.
9. If from a point in one of the assymptotes of an hyperbola, any straight line be drawn to intersect the curve (or opposite curves), in two points, and from the points of section, lines parallel to the same assymptote be drawn to meet the other, the sum (or difference) of the parallels will always be of the same magnitude.
The propositions in the exercises to the preceding book may be applied to the hyperbola.