Question 22 - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

Find the equation of the normal to the curve y = 𝑥 + 1/x, x > 0 perpendicular to the line 3𝑥 − 4𝑦 = 7.

Find the equation of the normal to the curve y = x + 1/x perpendicular

Question 22 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2

Question 22 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3

Question 22 - CBSE Class 12 Sample Paper for 2021 Boards - Part 4

Transcript

Question 22 Find the equation of the normal to the curve y = 𝑥 + 1/𝑥, x > 0 perpendicular to the line 3𝑥 − 4𝑦 = 7 Finding Slope y = x + 1/𝑥 Differentiating both sides 𝑑𝑦/𝑑𝑥 = 1 − 1/𝑥^2 Now, Slope of Normal = (−𝟏)/(𝟏 − 𝟏/𝒙^𝟐 ) Given that Normal is perpendicular to 3x − 4y = 7 So, Slope of Normal × Slope of Line = −1 (−1)/(1 − 1/𝑥^2 ) × 3/4 = −1 3/4 = 1 − 1/𝑥^2 1 − 1/𝑥^2 = 3/4 1 − 3/4 = 1/𝑥^2 1/4 = 1/𝑥^2 x2 = 4 x = ± 2 Since x > 0 ∴ x = 2 Finding y when x = 2 y = x + 1/𝑥 y = 2 + 1/2 y = 𝟓/𝟐 Now, Slope of Normal is (−𝟒)/𝟑 and it passes through point (2, 𝟓/𝟐) So, equation of Normal is (y − y1) = m (x − x1) y − 5/2 = − 4/3 (x − 2) y − 5/2 = − 4/3x + 8/3 Multiplying both sides by 6 6y − 6 × 5/2 = −6 × 4/3x + 6 × 8/3 6y − 15 = −8x + 16 8x + 6y = 31

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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