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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

Find the tangent and normal to the following curves at the given points on the curve y = x2 – x4 at (1, 0)

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प्रश्न

Find the tangent and normal to the following curves at the given points on the curve

y = x2 – x4 at (1, 0)

बेरीज
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उत्तर

y = x2 – x4 at (1, 0)

Differentiating w.r.t. ‘x’

`(("d"x)/("d"y))` = 2x – 4x3 

Slope of the tangent ‘m’ = `((dx)/("d"y))(1, 0)`

= 2(1) – 4(1)3

= – 2

Slope of the normal `- 1/"m" = (-1)/(-2) = 1/2`

Equation of tangent is

y – y1 = m(x – x1)

y – 0 = – 2(x – 1)

y = – 2x + 2

2x + y – 2 = 0

Equation of Normal is

y – y1 = `- 1/"m"` (x – x1)

y – 0 = `1/2(x - 1)`

2y = x – 1

x – 2y – 1 = 0

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पाठ 7: Applications of Differential Calculus - Exercise 7.2 [पृष्ठ १५]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 7 Applications of Differential Calculus
Exercise 7.2 | Q 5. (i) | पृष्ठ १५

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