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Class 12-commerce NCERT Solutions Maths Chapter 6 - Applications of Derivatives

Applications of Derivatives Exercise Ex. 6.1

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Applications of Derivatives Exercise Ex. 6.2

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Applications of Derivatives Exercise Ex. 6.3

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Applications of Derivatives Exercise Ex. 6.4

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Applications of Derivatives Exercise Ex. 6.5

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Applications of Derivatives Exercise Misc. Ex.

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OR

If we consider the equation of the curve as x2 = 4y

Differentiating with respect to x, we have:

2 x equals 4 fraction numerator d y over denominator d x end fraction rightwards double arrow fraction numerator d y over denominator d x end fraction equals x over 2 therefore right enclose fraction numerator d y over denominator d x end fraction end enclose subscript left parenthesis 1 comma 2 right parenthesis end subscript equals 1 half

Now, the slope of the normal at point (1, 2) is fraction numerator negative 1 over denominator right enclose fraction numerator d y over denominator d x end fraction end enclose subscript left parenthesis 1 comma 2 right parenthesis end subscript end fraction equals fraction numerator negative 1 over denominator 1 half end fraction minus 2

therefore Equation of the normal at (1, 2) is y minus 2 equals negative 2 left parenthesis x minus 1 right parenthesis

rightwards double arrow y minus 2 equals negative 2 x plus 2 rightwards double arrow 2 x plus y minus 4 equals 0

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