aa of the conjugate axis a line q=2, a third proportional to the con jugate and transverse axes. 2bb Since =p, we have bb=ap; substituting this value in the a two equations before found, we have yy=pr (aa-zz), according as the origin of the abscissæ is at one of the vertices, or at the centre. 30. A line FM or ƒ M drawn from the focus to any point of the ellipse is called a radius vector, if we suppose FC=c, we have b2 22 FM (yy+c2-2 cz+zz), or, since y-b and c-a-b2, b2 z2 as +a2—b2—2cz+xz)= fM -)—a—, and ƒM in the same manner will be by substitution FM= √ (b2— cz √(a2-2cz+ a2 a found equal to a + a Cz HenceƒM+FM=2 a=Aa, or the sum of the two lines drawn from the foci to any point of the ellipse, is always equal to the transverse axis; from this remarkable property we may deduce another method of describing an ellipse. Having fastened to two fixed points F and f a thread FMf, greater than Ff; extend this thread by means of a style M, and describe about the foci Ff a curve, which will be an ellipse, since the sum of the lines FM, ƒ M is every where the same. From this description it follows manifestly that on the same transverse axis we may describe an indefinite number of ellipses, some of which will approach nearer and nearer to the figure of the circumscribed circle; this will be the case of those whose two foci are the nearest together; while others will be more and more flattened in the direction of the conjugate axis in proportion as the foci recede from each other; so that the circle and straight line are the two limits of all ellipses. 31. Let it now be proposed to draw through the given point M, the tangent MT. Supposing the arc Mm infinitely small, draw from the foci F, ƒ, radius vectors fm, f'M, and Fm, FM; and describe from the centers F and f, and with the radii FM, the little arcs Mg, mr, and a M NF PAT we shall have fm+mF-FM+Mf-or ƒ M-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 ƒ 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 ƒMN will be equal to the angle NMF (because / OMF + 2ƒMN=2 TMF + 2 NMF, and we have just shewn that OMF, or its equal LMT=TMF) therefore we shall have ƒM: FM::ƒN: FN (Euclid), or by composition FM ;: ƒN+FN or, FN fM+FM or since c2=a2-b2, by substitution we have FN=c-z+ Consequently FN + z--c=PN= 62 z P. 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 (When the origin of the abscissæ is at the vertex PT: aa 2 ax-xx a-x Consequently CT-PT+CP=. = 4, 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 FmT, FMT may be considered as equal. The expression for the tangent MT mav be found by means of the right-angled triangle PMT⚫ of the diameter CN, and the parts CP are the abscissæ. Lastly, the parameter or latus rectum of any diameter, is a third proportional to that diameter and its conjugate. 34. From the extremities D and N draw the two ordinates NQ, DI to the transverse axis Aa, call CQ=z, DI=u; because of the similar triangles DIC, NQT, we shall have and therefore CQ (z): DI (u)::a: b; in the a very same manner we shall find CI: NQ::a: b. Consequently CQ: DICI: NQ; therefore the triangles DIC CNQ, are equal in surface. Hence Io, DI b1 CQ -bb-NQ', or DI'+NQ'=bb IIo, CIa = Œ NQ'=aa—CQ*, or CI*+CQ'=a* III°, a2+b=CI1+CQ'+NQ'+DI'=CD1+CN: that is, in the ellipse, the sum of the squares of any two conjugate diameters is always e ual to the sum of the squares of the two axes. IV°. If we draw ND, the surface of the triangle NCD will be expressed by ★ (DI+NQ) (CI+CQ)— CI × ID— b CQ × NQ = { CI× NQ+ b a ab. Hence the surface of the parallelogram CDEN will be equal to ab, and that of the whole parallelogram FEHG will be equal to 4 ab 2 ax2 b. Consequently all parallelograms circumscribed about an ellipse are equal to each other, and to the rectangle of the two axes. 35. Call now the semi-diameter CN=m, CD=n, the angle CPM-DCn P, we shall have Io, m2+n2=a2+b2; 2ly, ab=mn sin P, which is an expression for the surface of the parallelogram CDNE. These two expressions enable us to find immediately the two equal conjugate diameters of an ellipse; for then we have 2 m2 a2+b2, or m=±√ 1⁄2 (a2+b2), and sin P= 2 ab And since a2+b2° these quantities are always real, every ellipse must have two equal conjugate diameters. With regard to their position, it depends upon the value of CQ; now a2+b2 a2 CQ + NQ, or b2+ CQ2 = (a+b); therefore CQ= a √2, this value is independant of b, and shews that the ordinate NQ continued, will determine the equal conjugate diameters, in all ellipses, which have the axis Aa in common. 36. Let us now investigate the equation to the co-ordinates CP, PM, and call CP=z', PM=y', NT=q, NQ=r, QT=s, CQ=t,. If we draw PK, MO, perpendicular to the axis, and LP perpendicular to MO; by the similar triangles NQT, MLP, we shall find ML= , PL; and the two other similar triangles CPK, CNQ, give ry q PK, CK=; hence CO= m m m a1 m from the property of the ellipse we have MO1 a2-CO1. But `b2 q2 $3 + — ) y2+ q2 +—) z22—a3, and since 62 Observe now that when x=o, y=n, therefore the coefficient of y' is when on the contrary, y=o, then x'=m, hence the coefficient of z' a1 Consequently the equation becomes y = (m2) a result perfectly agreeing with the equation to the two axes.* From the above equation it follows that any diameter NCn, bisects the ordinates MPm, and consequently the whole ellipse. And also that every diameter Nn is bisected at the centre C; for at the points N and n we have z'm'; and therefore z= ± m. 37. Problem I. Given the two semi-axes a and b, to find two diameters which shall make a given angle, P=DCn. (See preceding figure.) We have m2+n2=a2+b2, mn= 2ab sin P' ab sin P therefore m2+n* + 2mn= 2.ab and m2+n2-2mn =a2+b2 therefore m+n=√(aa +b2+; 2ab sin P 5), and m—n= √ (a2+b2—_ 2ab Consequently m = ↓ √(a2+b2+ 2ab) + b √(a2+ba sin P and n = } √(a2+b2 + 2ab) — } √ √ (a2+b2 — 2ab sin sin P sin P 2ab sin p') P We have now only to determine the direction of one of the diameters, or the angle ACN which call C. sin (P—C): m :: sin P: CT= a2 sin (P-C) m sin P 1: m :: cos C: =3 aa CQ The triangle CNT gives m sin P sin (P-C) ; hence CQ= ; therefore in the right angled triangle CNQ we have a2 sin (P—C), which gives m2 cos C sin P=aa sin (P m sin P -C) a' sin P cos C-a2 sin C cos P, or. Without supposing z' or y'o, we may equally arrive at the final equation. CD2 :: QT2: CI :: NQ2: DI2, or q2 n2: s2 ; a2-t2 ;; r2 |