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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 Dec. 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.

 

 

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

    3. CBSE Class 12 Sample Paper for 2021 Boards

    CBSE Class 12 Sample Paper for 2020 Boards →
CBSE Class 12 Sample Paper for 2020 Boards →
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Davneet Singh

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 and Computer Science at Teachoo