 | De Volson Wood - Geometry, Analytic - 1890 - 372 pages
...equation. Changing Ir to — b1, and b' 2 to — b' ° gives 4o' 2 - 46' 2 = 4«2 - 46" ; (6) that is : For the hyperbola the difference of the squares of any two conjugate diameters, equals the difference of the tquares of the axes. Because the axes are conjugate, we must have, (Art.... | |
 | Charles Hamilton Ashton - Geometry, Analytic - 1900 - 290 pages
...+ 6'2 = (a2 + i2) *\ + (a2 + i2) ?jj, f/ 0 But since Pi is on the ellipse, + = 1, and a'2 a2 o2 2. In the hyperbola the difference of the squares of any two conjugate semi-diameters is equal to the difference of the squares of the semi-axes. 3. The product of the focal... | |
 | Maxime Bôcher - Geometry, Analytic - 1915 - 258 pages
...of the conjugate diameter, prove that * = 2 ** * =- * 37. Hence, prove that in an ellipse, the sum of the squares of any two conjugate diameters is equal to the sum of the squares of the major and minor axes ; and that the product of the focal radii drawn to any... | |
| |
Fig. 83,84. conjugate diameters is equal to the sum of the squares of the... 
