Answer :

It is given that ay2 = x3

Now, differentiating both sides with respect to x, we get




Then, the slope of the tangent to the given curve at (am2, am3) is



Then, slope of normal at (am2, am3)


=


Therefore, equation of the normal at (am2, am3) is given by:


y - am3 =


3my – 3am4 = -2x + 2am2


2x + 3my – am2(2 + 3m2) = 0


Therefore, equation of the normal at (am2, am3) is 2x + 3my – am2(2 + 3m2) = 0


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