PROPOSITION IX. THEOREM. -The square of an ordinate to any diameter, is equal to four times the product of the corresponding abscissa, by the distance from the vertex of that diameter to the focus. Let AD be a tangent to the parabola VAM at the point A; through A draw the diameter HAC, and through H any point of the curve, as B, draw BC parallel to AD; draw also AF to the focus; then will the square of BC be equal to 4AF XAC. Draw CE parallel, and EBG G K D V F M perpendicular to the directrix HK; and join BH, BF, HF. Also, produce CB to meet HF in L. Because the right-angled triangles FHK, HCL are similar, and AD is parallel to CL, we have HF: FK:: HC: HL :: AC: DL. HFXDL= FK × AC, Hence (Prop. I., B. II.), or But 2HFXDL=2FK × AC, or 4VFX AC. 2HFXDL=HL'-LF' (Prop. X., B. IV.) -HG' or CE'. Hence CE is equal to 4VFX AC. Also, because the triangles BCE, AFD are similar, we have CE CB DF: AF. Therefore CE': CB':: DF: AF2 (Prop. X., B. II.) :: VF : AF (Prop. VII., Cor. 2) :: 4VFX AC: 4AF XAC. But the two antecedents of this proportion have been proved to be equal; hence the consequents are equal, or BC2-4AF XAC. Therefore, the square of an ordinate, &c. Cor. In like manner it may be proved that the square of CM is equal to 4AF x AC. Hence BC is equal to CM; and since the same may be proved for any ordinate, it follows that every diameter bisects its double ordinates. PROPOSITION X. THEOREM. The parameter of any diameter, is equal to four times the distance from its vertex to the focus. Let BAD be a parabola, of which F is the focus, AC is any diameter, and BD its parameter; then is BD equal to four times AF. Draw the tangent AE; then, since AEFC is a parallelogram, AC is equal to EF, which is equal to AF (Prop. IV.). Now, by Prop. IX., BC2 is equal to 4AFX AC; that is, to 4AF. Hence E B BC is equal to twice AF, and BD is equal to four times AF. Therefore, the parameter of any diameter, &c. Cor. Hence the square of an ordinate to a diameter, is equal to the product of its parameter by the corresponding abscissa. PROPOSITION XI. THEOREM. If a cone be cut by a plane parallel to its side, the section is a parabola. Let ABGCD be a cone cut by a plane VDG parallel to the slant side AB; then will the section DVG be a parabola. A D Ε Ꮐ Let ABC be a plane section through the axis of the cone, and perpendicular to the plane VDG; then VE, which is their common section, will be parallel to AB. Let bgcd be a plane parallel to the base of the cone; the intersection of this plane with the cone will be a circle. Since the B plane ABC divides the cone into two equal parts, BC is a diameter of the circle BGCD, and be is a diameter of the circle bgcd. Let DEG, deg be the common sections of the plane VDG with the planes BGCD, bgcd respectively. Then DG is perpendicular to the plane ABC, and, consequently, to the lines VE, BC. For the same reason, dg is perpendicular to the two lines VE, bc. Now, since be is parallel to BE, and bB to eE, the figure bBEe is a parallelogram, and be is equal to BE. But because the triangles Vec, VEC are similar, we have : ec: EC: Ve: VE; and multiplying the first and second terms of this proportion by the equals be and BE, we have be xec: BEXEC:: Ve: VE. But since bc is a diameter of the circle bgcd, and de is perpendicular to bc (Prop. XXII., Cor., B. IV.), be Xec de'. For the same reason, BEXEC=DE'. B A Substituting these values of beXec and BEXEC in the preceding proportion, we have de: DE':: Ve: VE; that is, the squares of the ordinates are to each other as the corresponding abscissas; and hence the curve is a parabola, whose axis is VE (Prop. VIII., Cor. 1.). Hence the parabola is called a conic section, as mentioned on page 177. PROPOSITION XII. THEOREM. Every segment of a parabola is two thirds of its circumscribing rectangle. Let AVD be a segment of a parabola cut off by the straight line AD perpendicular to the axis; the area of AVD is two thirds of the circumscribing rectangle ABCD. Draw the line AE touching the parabola at A, and meeting the axis produced in E; and take a point H in the curve, so near to A that the E B K A M L D tangent and curve may be regarded as coinciding. Through H draw KL perpendicular, and MN parallel to the axis. Then the rectangle AL: rectangle AM :: AG×GL: AB× AN ::AG×GE: AB×AG because GL or NH: AN::GE: AG. But GE is equal to twice GV or AB (Prop. V.); hence that is, AL: AM:2:1; AL is double of AM. Hence the portion of the parabola included between two or dinates indefinitely near, is double the corresponding portion of the external space ABV. Therefore, since the same is true for every point of the curve, the whole space AVG is double the space ABV. Whence AVG is two thirds of ABVG; and the segment AVD is two thirds of the rectangle ABCD. Therefore, every segment, &c. ELLIPSE. Definitions. 1. AN ellipse is a plane curve, in which the sum of the distances of each point from two fixed points, is equal to a given line. 2. The two fixed points are called the foci. and if the point D moves about F in 3. The center is the middle point of the straight line joining the foci. D F 4. The eccentricity is the distance from the center to either focus. Thus, let ABA'B' be an ellipse, F and F the foci. Draw the line 5. A diameter is a straight line drawn through the center, and terminated both ways by the curve. Α' D' B B' 6. The extremities of a diameter are called its vertices. Thus, through C draw any straight line DD' terminated by the curve; DD' is a diameter of the ellipse; D and D' are its vertices. 7. The major axis is the diameter which passes through the foci; and its extremities are called the principal vertices. 8. The minor axis is the diameter which is perpendicular to the major axis. Thus, produce the line FF to meet the curve in A and A'; and through C draw BB' perpendicular to AA'; then is AA' the major axis, and BB' the minor axis. 9. A tangent is a straight line which meets the curve, but, being produced, does not cut it. |