## Find the equation of the normal to the curve y = ๐ฅ + 1/x, x > 0 perpendicular to the line 3๐ฅ − 4๐ฆ = 7.

Last updated at Oct. 27, 2020 by Teachoo

Transcript

Question 22 Find the equation of the normal to the curve y = ๐ฅ + 1/๐ฅ, x > 0 perpendicular to the line 3๐ฅ โ 4๐ฆ = 7 Finding Slope y = x + 1/๐ฅ Differentiating both sides ๐๐ฆ/๐๐ฅ = 1 โ 1/๐ฅ^2 Now, Slope of Normal = (โ๐)/(๐ โ ๐/๐^๐ ) Given that Normal is perpendicular to 3x โ 4y = 7 So, Slope of Normal ร Slope of Line = โ1 (โ1)/(1 โ 1/๐ฅ^2 ) ร 3/4 = โ1 3/4 = 1 โ 1/๐ฅ^2 1 โ 1/๐ฅ^2 = 3/4 1 โ 3/4 = 1/๐ฅ^2 1/4 = 1/๐ฅ^2 x2 = 4 x = ยฑ 2 Since x > 0 โด x = 2 Finding y when x = 2 y = x + 1/๐ฅ y = 2 + 1/2 y = ๐/๐ Now, Slope of Normal is (โ๐)/๐ and it passes through point (2, ๐/๐) So, equation of Normal is (y โ y1) = m (x โ x1) y โ 5/2 = โ 4/3 (x โ 2) y โ 5/2 = โ 4/3x + 8/3 Multiplying both sides by 6 6y โ 6 ร 5/2 = โ6 ร 4/3x + 6 ร 8/3 6y โ 15 = โ8x + 16 8x + 6y = 31

CBSE Class 12 Sample Paper for 2021 Boards

Question 1 (Choice 1)

Question 1 (Choice 2) Important

Question 2

Question 3 (Choice 1) Important

Question 3 (Choice 2) Important

Question 4

Question 5 โ Choice 1

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Question 7 (Choice 1)

Question 7 (Choice 2)

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Question 20 (Choice 1)

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Question 22 Important You are here

Question 23 (Choice 1)

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Question 24

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Question 26 Important

Question 27 Important

Question 28 (Choice 1)

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Question 31 (Choice 1)

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Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.