IIT JEE Advanced 2006 Paper 1

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.