1887.] A. Mukhop^dhjaj— Differential Equation of a Trajectory. 119 



expressed in a very neat and symmetrical form, in terms of a parameter 

 <p viz.f we have 



x = Ji cos </>. cosh n(\-\-<p). ) 



(A). 



y = h sin (p. sinh n(X + </>), 



§ 3. The equations (A) admit of a very simple geometrical inter- 

 pretation. 



Let A'A be the line joining the 

 foci of the system of confocal ellipses, 

 so that OA = h. On A'A as diameter, 

 describe a circle having its centre at O. 

 Draw any radius OB, making the angle 

 AOB = <p ; draw BO perpendicular to 

 OA ; then, we have OC = h cos <p, BO 

 = h sin </>. Construct a hyperbola 

 CM, having its centre at O, and its 

 transverse and congugate axes equal 

 to OC, BO respectively; then, of 

 course, BO is a tangent and OB an asymptote to this hyperbola. Take 

 a point M on the hyperbola, so that the area of the hyperbolic sector 

 OCM maybe w(X-i-</>) times the area of the triangle OCB : then, I assert 

 that M is a point on the trajectory, viz., the co-ordinates of M are 



x=h cos <p. cosh n(X-f </>). 

 y = h sin </>. sinh n(\-\-(p). 



To see how this is, drop MN perpendicular on OA. Then, writing 

 for the moment OC=h cos <^ = a, BO = /t sin <^ = )8, the hyperbola is 



2 ^2-^ 



and, any point («i, 2/i) ^^ M on this hyperbola is obviously satisfied by 

 iCj = a cosh \f/, 2/i =i^ sinh xj/, Now, the area of the portion CMN is given 



bv CMN= / ydx = !:: ^^x^-a^don 

 ^ •'a aJa 



= I- S aJv^a;* - a2 - a2 log (0^4- \/^^ - a^) > 



^a. K, ) x = a 



13 ( . , , . , , ^1 a cosh 1/^ + a sinh j/'"^ 

 = ^ < a^ posh \f/ smh i^ - a'* log > 



[•.• (Uj =a cosh i/'] 



= — j a* cosh 1^ sinh if/ — a^ if/ j 



