Q. 14 A3.6( 39 Votes )

Find the equations of the tangent and normal to the given curves at the indicated points:

y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)

Answer :

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


On differentiating with respect to x, we get


 = 4x- 18x2 +26x -10


Now, Slope of tangent will be the value of  at x = 0 i.e. 
m1 = 4(0)3 - 18(0)2 + 26(0) - 10 = -10

Therefore, the slope of the tangent at (0, 5) is -10.


Then, the equation of the tangent is

y – 5 = -10(x – 0)

y – 5 = 10x


10x +y =5



Also, slope of normal at (0,5)


Now, equation of the normal at (0,5)

10y - 50 = x

x -10y +50 = 0

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