Mathematical Problems on the First and Second Divisions of the Schedule of Subjects for the Cambridge Mathematical Tripos ExaminationMacmillan and Company, 1878 - 480 páginas |
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Resultados 1-5 de 40
Página 125
... moves along the straight line which is its locus . 753. Two diameters PP ' , QQ ' of a conic are drawn , and PR , PR ' let fall perpendicular on P'Q , P'Q ' ; prove that the chord intercepted by the conic on RR ' subtends a right angle ...
... moves along the straight line which is its locus . 753. Two diameters PP ' , QQ ' of a conic are drawn , and PR , PR ' let fall perpendicular on P'Q , P'Q ' ; prove that the chord intercepted by the conic on RR ' subtends a right angle ...
Página 136
... moves so that the straight lines joining it to two fixed points are equally inclined to a given direction : prove that its locus is a R. H. of which the two fixed points are ends of a diameter . 866. Circles are drawn through two given ...
... moves so that the straight lines joining it to two fixed points are equally inclined to a given direction : prove that its locus is a R. H. of which the two fixed points are ends of a diameter . 866. Circles are drawn through two given ...
Página 183
... move up to P ; and prove that its locus for different positions of P is where 4 { S + ( a - b ) U } +27 a2b2 ( a - b ) S = 0 , S = b2x2 - a2y ' - a'b ' , and U = x2 + y2 − a3 + b2 . - V. Polar Co - ordinates . 1122. The equation of the ...
... move up to P ; and prove that its locus for different positions of P is where 4 { S + ( a - b ) U } +27 a2b2 ( a - b ) S = 0 , S = b2x2 - a2y ' - a'b ' , and U = x2 + y2 − a3 + b2 . - V. Polar Co - ordinates . 1122. The equation of the ...
Página 191
... To reduce the equation u = 0 , if ab - h3 be finite first move the origin to the point whose co - ordinates satisfy the equations du • du 0 , 0 ; = d . dy the equation will become ax2 + by2 + 2hxy = CONIC SECTIONS , ANALYTICAL . 191.
... To reduce the equation u = 0 , if ab - h3 be finite first move the origin to the point whose co - ordinates satisfy the equations du • du 0 , 0 ; = d . dy the equation will become ax2 + by2 + 2hxy = CONIC SECTIONS , ANALYTICAL . 191.
Página 205
... Move the origin to the given point , and let the equation of the conic be u = ax2 + by2 + c + 2fy + 2gx + 2hxy = 0 , and let px + qy = 1 be the equation of a chord . The equation of the straight lines joining to the origin the ends of ...
... Move the origin to the given point , and let the equation of the conic be u = ax2 + by2 + c + 2fy + 2gx + 2hxy = 0 , and let px + qy = 1 be the equation of a chord . The equation of the straight lines joining to the origin the ends of ...
Otras ediciones - Ver todas
Mathematical Problems on the First and Second Divisions of the Schedule of ... Joseph Wolstenholme Sin vista previa disponible - 2012 |
Mathematical Problems: On the First and Second Divisions of the Schedule of ... Joseph Wolstenholme Sin vista previa disponible - 2017 |
Términos y frases comunes
angular points angular velocity asymptotes ax² axes bisected cardioid centre of perpendiculars chord circumscribed circle co-ordinates common point common tangents confocal conicoid conjugate diameters constant continued fraction cos² curve described diagonals directrix envelope equal excentric angles fixed circle fixed point fixed straight line foci focus four points given circle given conic given ellipse given point given the equations inscribed latus rectum length locus major axis meet minor axis normal parabola parallel particle passes plane point of intersection points of contact polar pole prove radical axis radius of curvature ratio rectangle rectangular hyperbola respectively right angles roots self-conjugate sin² sin³ straight line joining subtends a right tangents drawn tetrahedron triangle ABC velocity vertex vertical
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