Q. 14 B4.8( 16 Votes )

# Find the equation

Answer :

It is given that equation of curve is y = y = x4 – 6x3 + 13x2– 10x + 5

On differentiating with respect to x, we get = 4x3 - 18x2 +26x - 10 Therefore, the slope of the tangent at (1, 3) is 2.

Then, the equation of the tangent is

y – 3 = 2(x – 1)

y – 3 = 2x - 2

y = 2x +1

Then, slope of normal at (1,3)

= Now, equation of the normal at (1,3)

y – 3 = (x – 1)

2y -6 =- x + 1

x + 2y - 7 = 0

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