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analytical analytical representation asymptotes axis of abscissas becomes bisect called centre circumference coefficients conical surface conjugate diameters consequently construction cosine curve denote difference directrix distance draw ellipse equa equal equilateral expression find the equation formulas Geom geometrical given point given straight line gles hence hyperbola hyperboloid inclination indeterminate equation inscribed circle length Let ABC line passing locus major diameter negative oblique ordinate origin parabola perpendicular plane of xy plane triangle point of contact point of intersection positive primitive axes primitive equation principal diameters PROBLEM projections proposed line radii radius vector rectangle rectangular axes referred represent right angle secant second order sections semi-diameters square substituting subtangent suppose surface surface of revolution system of conjugate tangent tion transformed vertex vertical angle whence
Page 50 - In a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle, to determine the sides of the triangle.
Page 108 - In the first case, there must obviously subsist the conditions r -|- d~7 r, r -f- r' -7 d, r -|- d! ~7 r, which prove that if two circumferences cut, the distance of their centres must be less than the sum, and greater than the difference of the radii. In the second case where y becomes imaginary because of a negative factor, we must have one of the conditions d...
Page 201 - Given the base, an angle adjacent to the base, and the difference of the sides of a triangle, to construct it.
Page 30 - Having given the side of a regular decagon inscribed in a circle whose radius is known, to find the side of a regular pentagon inscribed in the same circle.
Page 163 - 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 semicircle...
Page 196 - Given the base of a triangle and the difference of the angles at the base, to determine the locus of the vertex. Taking the same axes as before, and putting a, a', for the tangents of the angles at the base...
Page 153 - ... does. From the general expression for the subtangent just given, it follows that T,x' = (x' + A') (x' — , A'), 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 120 - From these expressions for y and x, it appears that for the same value of x, there are two values of y numerically equal, but having contrary signs ; hence the chord AB bisects all the chords drawn parallel to CD. In like manner with regard to x ; this also has two values numerically equal, but differing in sign for one value of y; therefore, the chord CD bisects all the chords drawn parallel to AB. Moreover, since, for x = A, or x — — A, the corresponding value of y is 0, it follows that parallels...
Page 163 - 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 passes through thefocut.