If the tangent at any point P4m2,8m3 of x3−y2=0 is also a normal to curve x3−y2=0 , then m=

x3−y2=0……… (1)  Differentiating with respect to x2ydydx=3x2∴ Slope of tangent at P=dydx=3x22y4m2,8m3=3m∴ Equation of tangent at P is y−8m3=3mx−4m2 Or y=3mx−4m3…….. (2)  Solving (1) and (2)  (i.e) x3=y2 and y=3mx−4m33mx−4m32=x3x3−9m2x2+24m4x−16m6=0By trial and error method, x=m2 satisfies above equations.After further factorization we getx−4m22(x−m)2=0∴x=4m2,m2 Put x=m2 in (2); then y=−m3∴Qm2,−m3 Slope of tangent at Q=dydxm2,−m3=−32m Slope of normal at Q=23m∴  23m=3m (Or) 9m2=2∴m=±23 Therefore, the correct answer is (1).