The Analyst: A Monthly Journal of Pure and Applied Mathematics, Volumes 3-4Pierson & Blair, 1876 |
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Page 79
... hyperbolas whose semi axes are R and √ ( R2 — a — b3 ) . In this case we must have a > R in order to render the ... hyperbola for generatrix the section is imaginary , the equation of its projec- tion being R2x2 + ( R2 + a2 ) ( y ...
... hyperbolas whose semi axes are R and √ ( R2 — a — b3 ) . In this case we must have a > R in order to render the ... hyperbola for generatrix the section is imaginary , the equation of its projec- tion being R2x2 + ( R2 + a2 ) ( y ...
Page 97
... and position that one pierces entirely through the other , cutting from it two separate curves . These being projected on the plane of xy in a conic , the latter must be a hyperbola , since no other conic has two separate branches.
... and position that one pierces entirely through the other , cutting from it two separate curves . These being projected on the plane of xy in a conic , the latter must be a hyperbola , since no other conic has two separate branches.
Page 98
... hyperbola will approach each other , i . e . their transverse axis will become shorter , and when the two quadrics exactly cut each other off the hyperbola will be re- duced to a pair of intersecting right lines , that is the curves of ...
... hyperbola will approach each other , i . e . their transverse axis will become shorter , and when the two quadrics exactly cut each other off the hyperbola will be re- duced to a pair of intersecting right lines , that is the curves of ...
Page 23
... hyperbolas , passing through A and B , and having the transverse axis parallel to AB . If a = b = c , ( 1 ) becomes ( a - B ) × ( B — † ) ( 7 — α ) = 0 , which is the equation to the bisectors of the angles of the triangle . 134 ...
... hyperbolas , passing through A and B , and having the transverse axis parallel to AB . If a = b = c , ( 1 ) becomes ( a - B ) × ( B — † ) ( 7 — α ) = 0 , which is the equation to the bisectors of the angles of the triangle . 134 ...
Page 44
... hyperbola is taking k = a , the inverse is the lemniscate , • p * == ( x2 + y2 ) 2 = a2 ( x2 — y2 ) . .. ( 4 ) . ( 5 ) . ( 6 ) Again , in the case of the cissoid , if we transfer the origin to the point ( a , 0 ) [ centre of the ...
... hyperbola is taking k = a , the inverse is the lemniscate , • p * == ( x2 + y2 ) 2 = a2 ( x2 — y2 ) . .. ( 4 ) . ( 5 ) . ( 6 ) Again , in the case of the cissoid , if we transfer the origin to the point ( a , 0 ) [ centre of the ...
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Common terms and phrases
5th degree a₁ algebraic ANALYST angle ARTEMAS MARTIN assumed axis b₁ C₁ circular coefficients common logarithms constant curve denote determine differential distance E. B. SEITZ ellipse Elliptic Functions equal equation expression factor formula function given circle gives hence hyperbola integral intersection inverse Lagrange's Theorem locus logarithms method multiply observations orbital orbital force ORSON PRATT oxen parabola parallel perpendicular plane points at infinity position probable error quadrics quadrilateral radius ratio reduce represent result roots sides SOLUTION BY PROF SOLUTIONS of problems sphere square substitution surface tangent Taylor's Theorem Theorem tion triangle UNION SPRINGS unknown quantities values velocity whence zero