BITSAT 2017 Slot 2 Solved Paper

Shortest distance between two curves occured along the common normal.
∴ Normal to y2=4x at (m2,2m) is
y+mx−2m−m3=0
Normal to y2=2(x−3) at (

m2

2

+3,m) is
y+m(x−3)−m−

m3

2

=0
Both normals are same, if
−2m−m3=−4m−

1

2

m3⇒m=0,2
So, points will be (4,4) and (5,2) or (4,−4) and (5,−2).
Hence, shortest distance will be
√(1+4)=√5