 | John Bonnycastle - 1829 - 256 ページ
...II. The transverse and . conjugate diameters and an ordinate being given, to find the two abscissas. RULE. As the conjugate diameter is to the transverse, So is the square root of the sum of the squares of the ordinate and semi-conjugate, To the distance between the ordinate and the... | |
 | William Grier - 1832 - 366 ページ
...therefore, (160 X 40) i : 60 : : 200 : 150 = the conjugate axis. 4. To find the transverse axis. Take the square root of the difference of the squares of the ordinate and half conjugate, and add to this the half conjugate if the lesser abscissa is used, but subtract the... | |
 | Tobias Ostrander - 1833 - 172 ページ
...III. The transverse and conjugate diameters and an ordinate being given, to find the two abscissas. Rule —-As the conjugate diameter is to the transverse, so is the square root of the sum of the squares of the ordinate and semi-conjugate to the distance between the ordinate and the... | |
 | Tobias Ostrander - 1834 - 182 ページ
...III. The transverse and conjugate diameters and an ordinate being given, to find the two abscissas. Rule — As the conjugate diameter is to the transverse, so is the square root of the sum of the squares of the ordinate and semi-conjugate to the distance between the ordinate and the... | |
 | Ireland commissioners of nat. educ - 1834 - 370 ページ
...quantities, we get d — ^— = •VM —. =- | ; lionet ' -iy \ 4v v / which is the rule. For K(4 c1 — y2) is the square root of the difference of the squares of the semi-conjugate and ordinate, to which ^ c it added for one factor ; then from the nature of proportion... | |
 | William Grier - 1836 - 384 ページ
...160, then, v/(160 X 40) : 60 : 200 : 150 = the conjugate axis. 4. To find the transverse axis. Take the square root of the difference of the squares of the ordinate and half conjugate, and add to this the half conjugate if the lesser absciss is used, but subtract the... | |
 | Commissioners of National Education in Ireland - 1837 - 290 ページ
...conjugate, and ordinate, to find the abscissas. RULE. As the conjugate is to the transverse diameter, so is the square root of the difference of the squares...semi-conjugate, to the distance between the ordinate and centre. Then this distance being added to, and subtracted from, the semi-diameter, will give the two abscissas.*... | |
 | John Bonnycastle, Benjamin Hallowell - 1839 - 100 ページ
...14 : 10=the ordinale required. CASE п. — FACE 96. Ex. 2. Here <J 12.5a— 10»= v/56.25=7.5=the square root of the difference of the squares of the ordinate and semi-conjugate, And 25 : 35 :: 7.5 : 10.5=the distance between the ordinate and centre, Hence semi-transverse 17.5... | |
 | William Draper Swan - 1841 - 40 ページ
...a square number, what must be done ? How is a mean proportional between two numbers found ? To what is the square root of the difference of the squares of the hypotenuse and perpendicular, equal ? What is a diagonal 1 What is a circle ? What is a diameter ?... | |
 | Commissioners of National Education in Ireland - 1841 - 136 ページ
...have d* — -5- + - = -j- — с2 x . . /e4 ж2 с2 — У2)] which is the rule. For <J (£ c3 — y2) is the square root of the difference of the squares of the semi-conjugate and ordinate, to which £ с is added for one factor ; then, from the nature of proportion,... | |
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As the conjugate diameter is to the transverse, So is the square root of the difference of the squares of the ordinate and semi-conjugate, To the distance between the ordinate and centre. 
