We have the equations y = x2 and y = −(x2 – 4x + 4) y = − (x–2)2
The equation of tangent of these two parabolas is given by y = m ×
1
4
m2 (1) and y = m (x - 2) +
1
4
m2 (2) The identical line is
1
1
=
m
m
=
−(
1
4
)m2
(
1
4
)m2−2m
1
4
m2 = 2m -
1
4
m2
1
2
m2 = 2m ⇒ m2 = 4m ⇒ m = 0 , 4 Therefore, the equation of common tangent of y =x2 and y = −(x2 – 4x + 4) is, respectively, y = 0 and y = 4x – 4.