(1)
Since (h, k) lies on the parabola y = x2, we have k = h2 So (1) gives
D ≡ D(k) = 
2
(
)
k
k
c
+
−
or
D′(k) =
2
1
2(
)
2
(
)
k
c
k
k
c
+
−
+
−
Now
D′(k) = 0 gives 
2
1
c2
k
−
=
Observe that when 
2
1
c2
k
−
<
, then 2(
)
1
0
k
−c
+ <
, i e , D ( )
0
′k
<