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
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CBSE Class 12 Sample Paper for 2021 Boards
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