Normal to y2=8ax is . . . . (i) y=mx−4am−2am3 and normal to y2=4b(x−c) with slope m is y=m(x−c)−2bm−bm3....(ii) Since, both parabolas have a common normal. ∴4am+2am3=cm+2bm+bm3 ⇒4a+2am2=c+2b+bm2 or m=0 ⇒(4a−c−2b)=(b−2a)m2 or (X -axis is common normal always) Since, x -axis is a common normal. Hence all the options are correct for m=0.