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

Question

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

Infinity Learn · Accepted Answer

$$x3$$−$$y2=0$$……… (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).

$If the tangent at any point P (4 m^{2}, 8 m^{3}) of x^{3} - y^{2} = 0 is also a normal to curve$ $x^{3} - y^{2} = 0, then m =$

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

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$If the tangent at any point P (4 m^{2}, 8 m^{3}) of x^{3} - y^{2} = 0 is also a normal to curve$ $x^{3} - y^{2} = 0, then m =$