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

 

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  1. Class 12
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Transcript

Question 22 Find the equation of the normal to the curve y = ๐‘ฅ + 1/๐‘ฅ, x > 0 perpendicular to the line 3๐‘ฅ โˆ’ 4๐‘ฆ = 7 Finding Slope y = x + 1/๐‘ฅ Differentiating both sides ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1 โˆ’ 1/๐‘ฅ^2 Now, Slope of Normal = (โˆ’๐Ÿ)/(๐Ÿ โˆ’ ๐Ÿ/๐’™^๐Ÿ ) Given that Normal is perpendicular to 3x โˆ’ 4y = 7 So, Slope of Normal ร— Slope of Line = โˆ’1 (โˆ’1)/(1 โˆ’ 1/๐‘ฅ^2 ) ร— 3/4 = โˆ’1 3/4 = 1 โˆ’ 1/๐‘ฅ^2 1 โˆ’ 1/๐‘ฅ^2 = 3/4 1 โˆ’ 3/4 = 1/๐‘ฅ^2 1/4 = 1/๐‘ฅ^2 x2 = 4 x = ยฑ 2 Since x > 0 โˆด x = 2 Finding y when x = 2 y = x + 1/๐‘ฅ y = 2 + 1/2 y = ๐Ÿ“/๐Ÿ Now, Slope of Normal is (โˆ’๐Ÿ’)/๐Ÿ‘ and it passes through point (2, ๐Ÿ“/๐Ÿ) So, equation of Normal is (y โˆ’ y1) = m (x โˆ’ x1) y โˆ’ 5/2 = โˆ’ 4/3 (x โˆ’ 2) y โˆ’ 5/2 = โˆ’ 4/3x + 8/3 Multiplying both sides by 6 6y โˆ’ 6 ร— 5/2 = โˆ’6 ร— 4/3x + 6 ร— 8/3 6y โˆ’ 15 = โˆ’8x + 16 8x + 6y = 31

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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.