If an ordinate be drawn to any diameter of a parabola, the square of the semi-ordinate is equal to the rectangle under the abscissæ and the parameter.

Let A N be any diameter of the parabola to which PN Q is an ordinate. Draw the parameter TSV, cutting the diameter in F; join SA, and let the diameter meet the directrix in D. Because TF is half of TV, Prop. XXIII. and AF is half of DF or TF, Prop. XVII. Cor. 2, AF: ET:: FT: TV, and A FTV FT2; but AFTV: ANTV:: TF2: PN2, Prop. XXI., therefore AN-TV = P N2.

COR. Because TV2 TF4SA, 4 SAA N=PN2.

PROP. XXX. Fig. 38.

If two tangents be drawn at the extremities of any right line which is terminated by a conic section, and which does not pass through the centre of an ellipse, they will meet each other in the diameter which bisects that right line.

Let PQ meet the curve in P and Q; bisect PQ in N, and through N draw the diameter CNT. Through P draw the tangent PT, meeting the diameter in T, and join TQ which will touch the conic section in Q. For draw any other line DCB parallel to P N Q, meeting TP, TQ, in L and M. The triangles TN P, TCL, are similar, as also TNQ, TCM, therefore,

NP:CL:: TN:TC::QN: MC, and alternately NP NQ CL: CM, therefore CM CL, which is greater than CD or CB, therefore M is without the section, and the line TQ meets the curve only in one point Q.

Con. If two right lines which touch a conic section meet each other, a right line drawn through the point of concourse, bisecting the line which joins the points of contact, will be a dia

meter of the section.

PROP. XXXI. Fig. 39.

If a tangent to any point in the parabola meet a diameter, and an ordinate be drawn to that diameter from the point of contact, the segment of the diameter between the vertex and the tangent will be equal to the abscissa.

Let TP, which touches the parabola in any point P, meet the diameter N A in T, and draw the ordinate P N. NA, will be equal to A T. Draw the tangent AI meeting PT in I; join AP, and draw the diameter IG cutting AP in F and PN in G; then AFFP, Cor. Prop. XXIX. Therefore AI or NG PG; and AI is half of NP; but AI: NP::TA:TN, therefore TA is the half of TP.

PROP. XXXII. Fig. 40, 41.

If a tangent to any point in an ellipse, or an hyperbola, meet a diameter, and from the point of contact an ordinate be drawn to that diameter, the semidiameter will be a mean proportional between the segments of the diameter, which are intercepted between the centre and the ordinate, and between the centre and the tangent.

Let PT touch the ellipse or hyperbola in any

:

point P, and meet the diameter MA in T; draw PNQ an ordinate to the diameter MA, CN : CA: CA: CT. Through the vertices A, M draw the tangents A I, M L, meeting PT in I and L, take CO=CN. Then Cor. Prop. XXVI. IP: IA:: LP: LM, and alternately IP:: LP: IA:: LM, and because AI, NP, ML, are parallel, AN: NM:: TA: TM, and by AM: TA; and by taking the halves of the ancomp. fig. 41, or by div. fig. 40, ON: AN:: tecedents CN: AN:: CĂ: AT, and by comp. fig. 40, or by div. fig. 41, CA: CN:: CT: CA and by inversion CN: CA:: CA: CT.

SECT. II. OF THE DESCRIPTIONS OF THE CONIC SECTIONS; AND OF DRAWING TANGENTS TO THE CURVES.

PROP. XXXIII. PROBLEM. Fig. 3, 4, 5. Plate I. The focus, directrix, and determining ratio being given, to find any number of points in the conic section.

Let DX be the directrix and S the focus, draw SD perpendicular to DX, which produce indefinitely. Draw LST perpendicular to DS; and take S L and S T to S1) in the determining ratio. Then LST, Cor. 3, def. is the latus rectum. Join DL, DT, and produce them indefinitely. Take DXDS and join X S meeting DT in G, which Prop. II. will be parallel to DL in the parabola, it will meet it in some point g in the direction DL in the ellipse, and in the opposite direction in the hyperbola. Through G trix, meeting DL, DT, and the axis in K, G, A and g draw KA G, g M k, parallel to the direcand g, k M; the points A, M, will therefore be the vertices of the axis. Through any point N, in the axis, between A and M in the ellipse, and anywhere on the same side of A with S in the parabola, and anywhere except between A and M in the hyperbola, draw QNq parallel to the directrix, and from the centre S, with a radius equal to Q N, describe a circle cutting Qq in the points P, p; and join S P, Sp, which are each equal to QN; therefore P and p are points in the curve by the first proposition; and in this way may any number of points be found

PROP. XXXIV. PROBLEM. Fig. 42. Flate V.

Two unequal straight lines being given, which bisect each other at right angles, to describe an ellipse, of which the given lines shall be the axes.

Let AM, Bb, be the given lines of which A M is the greater. From the centre B, with a radius equal to AC, describe a circle meeting A M in S and H, which will be the foci, Prop. VII. Cor. 1. Take a string equal in length to AM, and fix the extremities of it at the points S and H, and by means of a pin let the string be stretched, and let the pin be carried round, till it return to the of which AM, Bb, are the axes, as is evident same point, the point M will describe an ellipse, from Prop. XIÍ.

PROP. XXXV. PROBLEM. Fig. 43.

Two straight lines being given, which bisect each other at right angles, to describe an hyperbola, of which these lines shall be the axes.

same reason EF EK, therefore, HIFK is á circle of which E is the centre.

COR. If, from any point D in an ellipse or hyperbola, a diameter D F be drawn, and a perpendicular DH to its conjugate EG, the radius of curvature at the point D is a third proportional to the perpendicular DH and the semiconjugate diameter EC.

For since DH: DC. DP: DO,

and DC: EC:: EG: P or DP

PROP. XLI. Fig. 48.

If a scalene cone ABDC be cut through the axis by a plane perpendicular to its base, making the triangle ABC, and from any point Lin of the triangle, so that the angle ALM ABC, the straight line A C, LM, be drawn in the plane and the cone be cut by another plane passing through ABC; the common section LPMQ of this LM, perpendicular to the triangle plane and the cone will be a circle.

Take any point N in the straight line LM; draw FNG parallel to CB; and let FPGQ

PROP. XXXIX. Fig. 47.

If a cone be cut by a plane passing through the vertex, the section will be a triangle.

Let ABGC be a cone of which AD is the axis; let GB be the common section of the base and cutting plane; join AB, AG. When the generating line comes to B and G, it will concide with AB, AG; they are, therefore, in the surface of the cone, and they are in the plane which passes through the points A, B, G; therefore, the triangle ABG is the common section of the cone and the plane which passes through the

vertex.

PROP. XL. Fig. 47.

If a cone be cut by a plane parallel to its base the section will be a circle, the centre of which is in the axis.

Let HFK be the section made by a plane parallel to the base of the cone, and let ACB, A DG, be the two sections of the cone, made by any two planes passing through the axis AD; let KH, EF, be the common sections of the plane HFK and the triangles AC B, ADG. Then, because the planes HFK, BGC, are parallel, EH EF will be parallel to DB, DG, and EH: DB:: EA: AD: EF: DG; but DB = DG, therefore, EHEF; and for the

FG; then the two planes FPGQ, L P M Q, being perpendicular to the plane ABC, their common section PNQ is perpendicular to FNG; therefore, PNN Q and PN2 FNNG, but the angle ALM ABC AGF, and the angles at N being vertical, the triangles F LN, MGN, are similar, and MN: NG:: FN: NL; therefore, the rectangle M N NLFN·NG= PN2, therefore, the section L P.MQ is a circle,

of which LM is a diameter. This section is called a subcontrary section.

PROP. XLII. Fig. 48, 49, 50.

If a cone be cut by a plane which does not pass through the vertex, and which is neither parallel to the base, nor to the plane of a subcontrary section; the common section of the plane and the surface of the cone will be an ellipse, a parabola, or an hyperbola, according as a plane passing through the vertex parallel to the cutting plane, falls without the cone, touches it, or falls within the cone.

the common section of a plane passing through Let ABDC be any cone; and let STV be its vertex and the plane of the base, which will fall without the base, will touch it, or will fall within it; let PMQ be a section made by a plane parallel to ASV, through O the centre of the base OT draw perpendicular to S V, meeting the circumference of the base in B and C; let a plane pass through A, B, and C, meeting the plane AS V in the line AT, the surface of the cone in A B, A C, and the plane of the section PMQ in LM; then LM will meet the side AB in M, and it will meet the other side A C, fig. 48, in L, within the cone, it will be parallel to it, fig. 49, and it will meet it in fig. 50, produced beyond the vertex in K.

Take any point N in the line LM; let FPGQ be a plane passing through N parallel to the base; and let FNG, PNQ, be the common sections of this plane and the planes A BC, PMQ; then PNQ will be the parallel to S V, and GF parallel to BT: and BT being perpendicular to SV, FNG is perpendicular to PNQ, therefore PN NQ, and PN EN NG. First, if the line STV be without the base, fig. 48, through M and L draw MHLK parallel to CB; then because the triangles LNF, LM H, are similar, also MNG, MLK,

LN: FN: LM: HM, and

NM: NG:: LM: LK; therefore LNN M: FN-NG or PN2 :: LM2: HMLK, which ratio is the same wherever the point N

taken; the section LPM Q is therefore an ellipse of which LM is a diameter and PN Q an ordinate.

Secondly, if the line STV, fig. 49, touch the circumference of the base in C; let DLE be the common section of the base and the plane PMQ, which is parallel to PN, and perpendicular to BLC; and BLLC DL2, therefore PN: DL2:: FN-NG: BL·LC (or because NG=LG) :: FN: BE; but the triangles MNF, MLB, being similar, FN: BL:: MN: ML, therefore PN2: DL:: MN: ML; and the section D ME is a parabola, of which ML is a diameter, and PNQ an ordinate.

Lastly, fig. 50, let the line STV fall within the base; through the vertex A draw A H parallel to GF; and because the triangles MNF, MHA, are similar, as also K NG, KHA,

MN: NF: MH: HA and

KN: NG:: KH: HA; therefore MNNK: FN-NG, or N P2:: M HH K: HA2 that is in a constant ratio, therefore the section DM E is a hyperbola, of which M K is a diameter and PNQ an ordinate.

SECT. V. OF THE AREAS OF THE CONIC
SECTIONS.

PROP. XLIII. Fig. 51.

If any ordinate and abscissa of a parabola be completed into a parallelogram; the area of the parabola, included between the ordinate and the curve, is to the parallelogram as 2 to 3.

=

Let A N be the abscissa, and P Q the ordinate; let the parallelogram PQCB be completed; and let AN be divided i to indefinitely small equal parts, of which ND is one; through D draw HI parallel to PQ, cutting the parabola in F and G, and through F draw K E parallel to NA; take KR KP, and draw R L parallel to K E. By Prop. XXI. IIFHG; PN2:: HP: NA, but because DN is indefinitely small, PQ or 2 PN may be taken for HG; and PK HF, aiso NA PB, therefore 2PK PN: PN2: HP: PB; hence 2 PK, or PR: PN:: HP : PB, and as the parallelograms RB, PD, are also equiangular, they are equal, and the parallelogram PD: KB: 2: 1, and the sum of all the parallelograms in APN is to the sum of all those in A P B in the same ratio of 2 to 1; but the sum of all the parallelograms in APN approaches indefinitely near to the curvilineal area APPN, when their breadths are continually dininished; and in like manner the sum of all the parallelograms in APB approaches to the curvi

ABSCISSA, 13. 23. ASYMPTOTE, 17. AXIS, conjugate, 11. AXIS, major, 9.

AXIS, minor, 11.

AXIS of a cone, 33.

linear area AFPB; therefore area AFPN: area AFPB:: 2:1, and the area PAQ is to the parallelogram PBC Q as 2 to 3.

PROP. XLIV. Fig. 52, 53.

If two ellipses or two hyperbolas have a common axis, and an ordinate be drawn through the same point in the axis to each of the curves; the areas included between the common abscissa, the ordinates, and the two curves, also the whole areas of the ellipses will be to each other as the conjugate axes

take any abscissa AN, which is not greater than Let A P, AQ, be two ellipses or two hyperbolas, half the axis of the ellipse, and draw the ordinates NP, NQ. Let the abscissa AN be divided into any GN, &c.; draw the ordinates ERI, FSK, number of equal parts, A E, EF, FG, GTL, and complete the parallelograms AR, AI, ES, EK, &c.; also draw Fi, Kk, Ll, parallel to AN. Then it is evident that the difference EK, FL, G Q, and the inscribed parallelograms between the circumscribed parallelograms AI, Ei, Fk, Gl, is equal to GQ; and if paralelo

grams be inscribed in the same manner in the figure APN, the difference between these and the circumscribed parallelograms would be equal to GP, therefore the difference between each series of parallelograms, and the areas AQN, APN, will be less than the parallelograms GQ, GP, and because GP: G Q :: N P : N Q and each parallelogram in the figure APN is to the corresponding parallelograms in the figure A QN in the same ratio, the sum of all those in APN is to the sum of all those in AQN as NP is to NQ, which is the same ratio with that of the conjugate axes. Conceive the breadths of the parallelograms to be now diminished, and their number increased, ad infinitum, and the parallelograms A P N, Á Q N, will be ultimately equal to the areas APN, AQN, for the parallelograms GQ, GP, will now vanish, therefore the areas A P N, AQ N, are to each other as their conjugate axes; and if the sections be ellipses their whole areas are to each other in the same ratio.

COR. 1. If a circle be described about an ellipse, the area of the circle is to the area of the ellipse as the transverse axis is to the conjugate.

COR. 2. The area of an ellipse is equal to that of a circle whose diameter is a mean proportional between the two axes.

COR. 3. The areas of two ellipses are to each other as the rectangles under their axes.