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Tamil Nadu Board of Secondary EducationHSC Science Class 12

Find the points on curve y = x3 – 6x2 + x + 3 where the normal is parallel to the line x + y = 1729 - Mathematics

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Question

Find the points on curve y = x3 – 6x2 + x + 3 where the normal is parallel to the line x + y = 1729

Sum

Solution

y = x3 – 6x2 + x + 3

Differentiating w.r.t. ‘x’

Slope of the tangent `("d"y)/("d"x)` = 3x2 – 12x + 1

Slope of the normal = `1/(3x^2 - 12x + 1)`

Given line is x + y = 1729

Slope of the line is – 1

Since the normal is parallel to the line, their slopes are equal.

`1/(3x^2 - 12x + 1)` = – 1

3x2 – 12x + 1 = 1

3x2 – 12x = 0

3x(x – 4) = 0

x = 0, 4

When x = 0, y = (0)3 – 6(0)2 + 0 + 3 = 3

When x = 4, y = (4)3 – 6(4)2 + 4 + 3

= 64 – 96 + 4 + 3

= – 25

∴ The points on the curve are (0, 3) and (4, – 25).

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Chapter 7: Applications of Differential Calculus - Exercise 7.2 [Page 14]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Applications of Differential Calculus
Exercise 7.2 | Q 3 | Page 14

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