Answer :

It is given that equation of curve is y = x3


On differentiating with respect to x, we get


= 3x2



Therefore, the slope of the tangent at (1, 1) is 3.


Then, the equation of the tangent is


y – 1 = 3(x – 1)


y = 3x - 2


Then, slope of normal at (1,1)


=


Now, equation of the normal at (1,1)


y – 1 = (x – 1)


3y -3 =- x + 1


x + 3y - 4 = 0


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