The Elements of Analytical Geometry: Comprehending the Doctrine of the Conic Sections, and the General Theory of Curves and Surfaces of the Second OrderJohn Souter, School Library, 73, St. Paul's Church Yard; and J. and J.J. Deighton, Cambridge., 1830 - Conic sections - 312 pages |
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Common terms and phrases
A² B² A² y² algebraical analytical analytical representation asymptotes axis of abscissas axis of y B²x² becomes bisects called centre circle circumference coefficient conjugate diameters construct coordinates cos.² curve denote determine directrix distance ellipse equa equal equilateral hyperbola expression find the equation focus follows formulas Geom geometrical given point hence hyperbola imaginary curve inclination indeterminate equation latus rectum length line passing locus major diameter My² negative obviously ordinate parabola parallelogram perpendicular plane point of contact points of intersection positive primitive axes primitive equation principal diameters PROBLEM proposed line radii radius vector rectangle rectangular axes represent right angle secant second degree semi-diameter sides sin.² square substituting subtangent subtracting supplemental chords suppose surface system of conjugate tion transformed equation transverse axis triangle trigonometrical tangent vertex x²²
Popular passages
Page 49 - They may cut each other, having two points common, when the distance between the centers is less than the sum and greater than the difference of the radii.
Page 115 - In article (100), it was found that the principal parameter is equal to 4m, the coefficient of x, in the equation of the curve, when referred to rectangular conjugates. By analogy, the coefficient 4r, when any system of conjugates are employed, is called the parameter of that diameter, which is taken for the axis of...
Page 118 - FP • FA = FR • FA. From this property we may derive an easy method of drawing a tangent to a parabola from a point either within or without the curve. Thus, let P be a point either within or without the curve, through which it is required to draw a tangent. Draw PF, upon which describe a...
Page 104 - ... does. From the general expression for the subtangent, just given, it follows that that is, as in the ellipse, the rectangle of the subtangent and abscissa of the point of contact is equal to the rectangle of the sum and difference of the same abscissa and semi-transverse axis. Thus OM-MR = A'M-MB...
Page 84 - FT +PF, PF = PG, also NF = NG, consequently the triangles NFP, NGP, are equal, therefore the angles NPF, NPG are equal, as also their supplements FPR, GPR, hence NPR is a tangent to the ellipse. As the distance...
Page 156 - Given the base, an angle adjacent to the base, and the difference of the sides of a triangle, to construct it.
Page 10 - PROBLEM XI. From any point within an equilateral triangle perpendiculars are drawn to the three sides to find the sum, s, of these perpendiculars. Ans. s — alt. of the triangle. PROBLEM XII. Given the diagonal of a rectangle, 10, and its perimeter, 28, to find the sides. Ans. 6 and 8. PROBLEM XIII. PROBLEM XIV. Within a given triangle, S, is inscribed another, by joining the middle of its sides, ^nd within this second triangle is inscribed another, by similar means.
Page 59 - A2 (A2 — c2) (5). In this equation, x and y are the coordinates of any point in the curve, and the other terms are all constant ; this, therefore, is the equation of the curve. (42.) Let us now inquire at what points the. curve cuts the axes. For this purpose, put y — 0, in equation (5), and there results for the abscissa of the point where the curve cuts Ox, x—±A..
Page 119 - D' is a right angle, and the angle DPF = DPD', and PF = PD', .,. DFP is a right angle. In like manner, DFP' is a right angle ; hence, first, the part of the tangent intercepted between the point of contact and the directrix, subtends a right angle at the focus ; second, the line joining the points of contact of perpendicular tangents always poises through the focvi.
Page 49 - P, constant, and equal to two right angles, minus half the sum of the angles at the base of the proposed triangle, that is, to one right angle plus half the given angle ; hence the required locus is an arc described on AB, containing this angle. A similar arc below AB, belongs also to the locus. PROBLEM IX. (35.) The base and vertical angle of a triangle being given to find the locus of the intersection of the straight lines, drawn from the angles to the middle of the opposite sides. By art. (19,)...