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 D F' C F from the foci, is equal to a given line. Now, when the point 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'. C 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 C; 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, 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 B' 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 BF-FC (Prop. XI., B. IV.), which is equal to AC-FC' (Prop. IV.). Hence (Prop. X., B. IV.), BC (AC+FC) x (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 AC'BC'. 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 F' 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 par allel 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 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. Α' H A F' F K D E Draw FG parallel to EE' or But the angles V.); hence the angles DGF, DFG are equal to each other, and DG is equa. to DF. Also, because CH is parallel to FG, and CF is equal to CF; therefore HG must be equal to HF'. Hence FD+F'D is equal to 2DG+2GH or 2DH. But FD+F'D is equal to 2AC. Therefore 2AC is equal to 2DH, or AC is equal to DH. Also, the angle DHK is equal to DKH; and hence DK is equal to DH or AC. Therefore, if from the vertex, &c. 13 PROPOSITION VIII. THEOREM. Perpendiculars drawn from the foci upon a tangent to the ellipse, meet the tangent in the circumference of a circle, whose diameter is the major axis. Let TT be a tangent to the ellipse at D, and from F draw F/E perpendicular to T/T; the point E will be in the circumference of a circle described upon AA' as a diameter. Join CE, FD, F/D, and produce F/E to meet FD produced in G. Then, in the two triangles DEF', DEG, because DE is common to both triangles, the angles A T F' E H T at E are equal, being right angles; also, the angle EDF' is equal to FDT (Prop. V.), which is equal to the vertical angle EDG; therefore DF is equal to DG, and EF' is equal to EG. Also, because F/E is equal to EG, and F/C is equal to CF, CE must be parallel to FG, and, consequently, equal to half of FG. But, since DG has been proved equal to DF, FG is equal to FD+DF', which is equal to AA'. Hence CE is equal to half of AA' or AC; and a circle described with C as a center, and radius CA, will pass through the point E. The same may be proved of a perpendicular let fall upon TT' from the focus F. Therefore, perpendiculars, &c. Cor. CE is parallel to DF, and if CH be joined, CH will be parallel to DF". |