Manipal Entrance Test 2013 Math Paper

Centre and radius of the given circle is P(6,0) and √5, respectively. Now minimum distance between two curves always Occurs along a line which normal to both the curves. Equation of normal to y2=4x at (t2,2t) is
y=−tx+2t+t3
If it is normal to circle also, then it must pass though (6, 0).
∴ 0=t3−4t⇒t=0
or t=±2
⇒ A(4,4) and C(4,−4)
⇒ PA=PC=√20=2√5
⇒ Required minimum distance =2√5−√5=√5