26 KEPLEK'S PROBLEM. 



an equation of the sixth order, very complicated and difficult.* 

 But, if the given point is in the centre or punctum duplex of 

 the curve, the equation is a cubic one, wanting the second 

 term, and of course, easily resolved. 



It often happens, too, that the problem may be resolved. 

 in genei'al, for a curve ; but that, in one particular part of 

 the axis, the solution becomes impossible. As this is rather 

 a singular circumstance, we shall attend a little more minutely 

 to it. 



Let it be required to resolve the problem for the case of 

 comets, supposing those bodies to move in parobolic orbits. 



The general equation for x becomes x*/x-\-3c\fx = r x 



fj ft 



mD 2 



- : a cubic wanting the second term, and easily resolved. 



m + n 



But, in certain cases, viz., when c, the distance of the given 



point from the vertex, is less than 3 D x A*/-, -, - rs the 



'V 4 a (m + nf 



problem cannot be resolved; for, in this case, the cube of 

 one-third of the co-efficient of x is less than the square of half 

 the last term, which is the well-known irreducible case of 



* The equation is of the following form, a being the lemniscata's semi- 

 diameter : 



(l - a) a? 5 I 

 )-'J 12m 



~ a } - 



+ 3 a 4 - 9c 2 a 2 (1 + 2 - 2 



+ 6ca 4 (l - a)x 



a cubocubic having all its terms (a- 6 + A a; 5 + B x* + C a; 3 + D x* + E x + 

 F = 0), in which A, C, and E vanish when the centre of motion (or of the 

 radii vectores) is in the punctum duplex, and then the equation to x is X s -(- 

 B x* + Dx 2 + F = 0, reducible to the cubic z 3 + A z -f <f> = 0. So that 



(A 

 



is less than [-?-) the irreducible case of Cardan's rule. 



