10. An ordinate to a diameter, is a straight line drawn. from any point of the curve to the diameter, parallel to the tangent at one of its vertices. Thus, let DD' be any diameter, and TT' a tangent to the ellipse at D. From any point G of the curve draw GKG' parallel to TT/ and cutting DD' in K; then is GK an ordinate to the diameter DD'. Ε' Τ G' Ꭲ K ส G D It is proved in Prop. XIX., Cor. 1, that GK is equal to G'K; hence the entire line GG' is called a double ordinate. 11. The parts into which a diameter is divided by an ordinate, are called abscissas. Thus, DK and D'K are the abscissas of the diameter DD' corresponding to the ordinate GK. 12. Two diameters are conjugate to one another, when each is parallel to the ordinates of the other. Thus, draw the diameter EE' parallel to GK, an ordinate to the diameter DD', in which case it will, of course, be parallel to the tangent TT'; then is the diameter EE' conjugate to DD'. 13. The latus rectum is the double ordinate to the major axis which passes through one of the foci. Thus, through the focus F/ draw LL a double ordinate to the major axis, it will be the latus rectum of the ellipse. 14. A subtangent is that part of the axis produced which is included between a tangent and the ordinate drawn from the point of contact. T L D L F T Thus, if TT be a tangent to tre curve at D, and DG an ordinate to the major axis, then GT is the corresponding subtangent. PROPOSITION I. PROBLEM. To describe an ellipse. Let F and F be any two fixed points. Take a thread longer than the distance FF', and fasten one of its extremities at F, the other at F'. Then let a pencil be made to glide along the thread so as to keep it always stretched; the curve described by the point of the pencil will be an ellipse. For, in every position of the pencil, the sum of the distances DF, DF' will be the same, viz., equal to the entire length of the string. PROPOSITION II. THEOREM. The sum of the two lines drawn from any point of an ellipse to the foci, is equal to the major axis. Let ADA' be an ellipse, of which F, F are the foci, AA' is the major axis, and D any point of the curve; then will DF+DF' be A equal to AA'. For, by Def. 1, the sum of the distances of any point of the curve F' D F from the foci, is equal to a given line. Now, when the poinɩ D arrives at A, FA+F/A or 2AF+FF' is equal to the given line. And when D is at A', FA/+F'A' or 2A/F/+FF is equal to the same line. Hence 2AF+FF'=2A/F'+FF'; consequently, AF is equal to A/F'. Hence DF+DF', which is equal to AF+AF, must be equal to AA'. Therefore, the sum of the two lines, &c. Cor. The major axis is bisected in the center. For, by Def. 3, CF is equal to CF; and we have just proved that AF is equal to A/F/: therefore AC is equal to A/C. PROPOSITION III. THEOREM. Every diameter is bisected in the center. Let D be any point of an ellipse; join DF, DF', and FF. Complete the parallelogram DFD'F', and join DD'. F D F Now, because the opposite sides of a parallelogram are equal, the sum of DF and DF is equal to the sum of D'F and D'F; hence D' is a point in the ellipse. But the diagonals of a parallelogram bisect each other; therefore FF' is bisected in Ĉ; that is, C is the center of the ellipse, and DD' is a diameter bisected in C. Therefore, every diameter, &c. PROPOSITION IV. THEOREM. The distance from either focus to the extremity of the minor axis, is equal to half the major axis. Let F and F be the foci of an ellipse, AA' the major axis, and BB the minor axis; draw the straight lines BF, BF'; then BF, A' BF are each equal to AC. In the two right-angled triangles BCF, BCF', CF is equal to CF', and BC is common to both F' B Β ́ A F triangles; hence BF is equal to BF'. But BF+BF' is equal to 2AC (Prop. II.); consequently, BF and BF' are each equal to AC. Therefore, the distance, &c. Cor. 1. Half the minor axis is a mean proportional between the distances from either focus to the principal vertices. For BC is equal to BF2-FC' (Prop. XI., B. IV.), which is equal to AC-FC' (Prop. IV.). Hence (Prop. X., B. IV.), BC2=(AC+FC) × (AC-FC) =AF/XAF; and, therefore, AF: BC: BC: FA!. Cor. 2. The square of the eccentricity is equal to the difference of the squares of the semi-axes. For FC is equal to BF-BC, which is equal to AC3. BC2. PROPOSITION V. THEOREM. A tangent to the ellipse makes equal angles with straight 'ines drawn from the point of contact to the foci. Let F, F' be the foci of an ellipse, and D any point of the curve; if through the point D the line TT' be drawn, making the angle TDF equal to T'DF', then will TT be a tangent to the ellipse at D. For if TT be not a tangent, it must meet the curve in some other Γ ́ T' E G T F point than D. Suppose it to meet the curve in the point E. Produce F'D to G, making DG equal to DF; and join EF, EF, EG, and FG. Now, in the two triangles DFH, DGH, because DF is equal to DG, DH is common to both triangles, and the angle FDH is, by supposition, equal to F'DT', which is equal to the vertical angle GDH; therefore HF is equal to HG, and the angle DHF is equal to the angle DHG. Hence the line TT/ is perpendicular to FG at its middle point; and, therefore, EF is equal to EG. Also, F/G is equal to F'D+DF, or F'E+EF, from the nature of the ellipse. But F'E+EG is greater than F/G (Prop. VIII., B. I.); it is, therefore, greater than F/E+EF. Consequently ÉG is greater than EF; which is impossible, for we have just proved EG equal to EF. Therefore E is not a point of the curve, and TT' can not meet the curve in any other point than D; hence it is a tangent to the curve at the point D. Therefore, a tangent to the ellipse, &c. Cor. 1. The tangents at the vertices of the axes, are perpendicular to the axes; and hence an ordinate to either axis is perpendicular to that axis. Cor. 2. If TT represent a plane mirror, a ray of light proceeding from F in the direction FD, would be reflected in the direction DF', making the angle of reflection equal to the angle of incidence. And, since the ellipse may be regarded as coinciding with a tangent at the point of contact, if rays of light proceed from one focus of a concave ellipsoidal mirror, they will all be reflected to the other focus. For this reason, the points F, F' are called the foci, or burning points. PROPOSITION VI. THEOREM. Tangents to the ellipse at the vertices of a diameter, are parallel to each other. Let DD be any diameter of an ellipse, and TT' VV tangents to the curve at the points D, D'; then will they be parallel to each other. Join DF, DF, D'F, D'F'; then, by the preceding Proposition, the angle FDT is equal to F/DT', and the an V F T T gle FD/V is equal to F'D'V'. But, by Prop. III., DFD'F' is a parallelogram; and since the opposite angles of a parallelogram are equal, the angle FDF is equal to FD/F; therefore the angle FDT is equal to F'D'V' (Prop. II., B. I.). Also, since FD is parallel to F'D', the angle FDD' is equal to F/D/D; hence the whole angle D'DT is equal to DD/V'; and, consequently, TT' is parallel to VV'. Therefore, tangents, &c. Cor. If tangents are drawn through the vertices of any two diameters, they will form a parallelogram circumscribing the ellipse. PROPOSITION VII. THEOREM. If from the vertex of any diameter, straight lines are drawn through the foci, meeting the conjugate diameter, the part intercepted by the conjugate is equal to half the major axis. Let EE be a diameter conjugate to DD', and let the lines DF, DF' be drawn, and produced, if necessary, so as to meet EE' in H and K; then will DH or DK be equal to AC. T E G H Α' AT F' K D Draw FG parallel to EE' or E But the angles V.); hence the |