Q. 14 C3.7( 9 Votes )

# Find the equation

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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