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Find the equation of the normal to the curve y = π‘₯ + 1/x, Β  x > 0 perpendicular to the line 3π‘₯ βˆ’ 4𝑦 = 7.

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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

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Question 22 Find the equation of the normal to the curve y = π‘₯ + 1/π‘₯, x > 0 perpendicular to the line 3π‘₯ βˆ’ 4𝑦 = 7 Finding Slope y = x + 1/π‘₯ Differentiating both sides 𝑑𝑦/𝑑π‘₯ = 1 βˆ’ 1/π‘₯^2 Now, Slope of Normal = (βˆ’πŸ)/(𝟏 βˆ’ 𝟏/𝒙^𝟐 ) Given that Normal is perpendicular to 3x βˆ’ 4y = 7 So, Slope of Normal Γ— Slope of Line = βˆ’1 (βˆ’1)/(1 βˆ’ 1/π‘₯^2 ) Γ— 3/4 = βˆ’1 3/4 = 1 βˆ’ 1/π‘₯^2 1 βˆ’ 1/π‘₯^2 = 3/4 1 βˆ’ 3/4 = 1/π‘₯^2 1/4 = 1/π‘₯^2 x2 = 4 x = Β± 2 Since x > 0 ∴ x = 2 Finding y when x = 2 y = x + 1/π‘₯ y = 2 + 1/2 y = πŸ“/𝟐 Now, Slope of Normal is (βˆ’πŸ’)/πŸ‘ and it passes through point (2, πŸ“/𝟐) So, equation of Normal is (y βˆ’ y1) = m (x βˆ’ x1) y βˆ’ 5/2 = βˆ’ 4/3 (x βˆ’ 2) y βˆ’ 5/2 = βˆ’ 4/3x + 8/3 Multiplying both sides by 6 6y βˆ’ 6 Γ— 5/2 = βˆ’6 Γ— 4/3x + 6 Γ— 8/3 6y βˆ’ 15 = βˆ’8x + 16 8x + 6y = 31

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