162J THE GEOMETRY OF THE CYLINDROID. 157 



We can also obtain X ; for let it be Px+ Qy, then severally identifying the 

 coefficients of a? and y 3 , we have, 



P = m sin 2a 4- h tan ft, 

 Q = m cos ft (cos 2a + cos 20) ; 

 finally, resuming the various results, we obtain the identity 



wherein, 



c = - h 2 tan ft + m 2 cot /3 sin 2 20, 

 U = sin/3 cos 2 /3?/ 3 + sin Pyx?, 

 (m sin 2a + A tan /3) a? 2 + 2mxy cos /3 cos 2a 

 + (m sin 2a cos 2 ft h sin ft cos /3) t/ 2 . 



F= wz# 2 (cos 2a + cos 2$) + 2//m/ sin 2a cos ft + my 2 cos 2 ft (cos 20 cos 2a), 

 X = oc(m sin 2a + A tan /3) my cos ft (cos 2a + cos 20), 

 i = TO# (cos 2a + cos 20) + y (h sin /3 + m cos /3 sin 2a) 



- A 2 tan /3 + m 2 cot sin 2 20, 

 F= a? ( m sin 2a A tan /3) -f 2?n#y cos ^ cos 2a 



+ 2/ 2 (m sin 2a cos 2 /3 h sin y3 cos y3). 



162. Parabola. 



The screws reciprocal to a cylindroid intersect two screws of equal pitch 

 on the surface. Any chord in the section which cuts the cubic in two points 

 of equal pitch must thus be the residence of a screw reciprocal to the surface; 

 accordingly the chord 



mx (cos 2a + cos 20) + y (h sin ft + m cos ft sin 2a) 



- h- tan ft + m? cot ft sin 2 20 = 0, 

 when it receives a pitch equal to 



p Q m cos 20, 

 forms a screw reciprocal to the cylindroid. 



It is easily shown that the envelope of this chord is a parabola; differ 

 entiating with respect to we have 



x = 2m cot ft cos 20. 

 Eliminating we obtain 



x z + 4&amp;lt;mx cot ft cos 2&amp;lt;z + 4y (h cos ft + m cos ft cot ft sin 2a) - 4h? + 4m 2 cot 2 ft = 0. 

 The vertex of the parabola is at the point 



x = - 2m cot ft cos 2a ; y = h sec ft m cosec ft sin 2a. 



