188 
CALCULUS 
(a) 
x 2 y 2 
+ V = 
(P) 
it db — 
It is more convenient to retain both parameters, rather than to 
eliminate, but we must be careful to remember that only one is 
independent. If we choose a as that one, a = a, 
and differentiate with respect to a, we have : 
2x 2 2y 2 db n A . db' 
a 3 b 3 da \ da 
and hence 
0, 
Fig. 43 
( c ) 
X i _ y z 
a?~W-' 
Between (a), (b), and (c) we can eliminate a and b and thus get a 
single equation in x and y, which will be the equation of the en 
velope. To do this, solve (a) and (c) for a 2 and b 2 , thus getting 
a 2 — 2 x 2 , b 2 = 2 y 2 , 
and then substitute the values of a and b from these equations in (b): 
± 2-n-xy — k, 
This equation represents a pair of equal equilateral hyperbolas on 
the axes as asymptotes. 
The equations 
x = ± a/V2, y = ± 6/V2, 
combined with (6), give the coordinates of the points of the en 
velope in which the particular ellipse corresponding to that pair of 
values of a and b is tangent to it. This remark applies generally 
whenever the coordinates x and y of a point of the envelope are 
obtained as functions of a. 
EXERCISES 
1. Find the envelope of the family of straight lines 
2 ay = 2 x -f- a 2 . 
Draw a number of the lines. 
2. The same question for the family 
x cos a + y sin a = 2. 
3- The legs of a right triangle lie along two fixed lines, and the