88 DIFFERENTIAL AND INTEGRAL CALCULUS. 



therefore 

 AB=AM+MC + CB= 



d 2 4m a (a -f 



+ a = v - 



4wi 4m 



therefore 



hence, area of parabola cut off by DD' 



3m 

 to be made a minimum by variation of m). 



9. Find the minimum chord which can be drawn in 

 a given ellipse normal to the ellipse. 



[If 2a, 2b be the axes of the ellipse, a cos#, b sin0 the 

 point at which the chord is normal, and 2u its length, then 



u = 

 and 



]_du_, _ 

 ud0~ (a 



(d*-V} sin(9 cos^{o*(a it -25*) sin 2 6 - b* (2o' - ' J ) cos 2 6>} _ 

 (a 2 sin 2 + 6 2 cos' 2 0) (a 4 sin 2 + b* cos" 0) 



and, therefore, -^ changes sign when = 0, 6 = a, = |?r, 



5 //2* - ft"\ 

 6 = 7r a, where tana = - . / ( ~ 972)? an " since when 



is a small negative quantity -^ is positive, ^ = gives a 



maximum, Q = a. a minimum, # = JTT a maximum, O = TT a. 

 a minimum, = TT a maximum (same as 6 = 0), and so on. 

 This assumes that 2 >2Z> 2 , otherwise tana is impossible; 



and if a s <26 5! , -^ changes sign only when 6=0 and = JTT, 



or the major axis is the only maximum and the minor axis 

 the only minimum normal chord. See fig. (18), where PQ 



