


Tangents and Normals (using Differentiation)
Tangents and Normals (using Differentiation)
Last updated at Dec. 16, 2024 by Teachoo
Question 10 Find the equation of all lines having slope –1 that are tangents to the curve 𝑦=1/(𝑥 − 1) , 𝑥≠1.Equation of Curve is 𝑦=1/(𝑥 − 1) Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(1/(𝑥 − 1))/𝑑𝑥 (𝑑𝑦 )/𝑑𝑥 =(𝑑 )/𝑑𝑥 (𝑥−1)^(−1) (𝑑𝑦 )/𝑑𝑥 =−1〖 × (𝑥−1)〗^(−1−1) (𝑑𝑦 )/𝑑𝑥 =−(𝑥−1)^(−2) 𝑑𝑦/𝑑𝑥=(− 1)/(𝑥 − 1)^2 Given that slope = −1 Hence, 𝑑𝑦/𝑑𝑥 = −1 ∴ (− 1)/(𝑥 − 1)^2 =−1 1/(𝑥 − 1)^2 =1 1=(𝑥 − 1)^2 (𝑥 − 1)^2=1 𝑥 − 1=±1 x − 1 = 1 x = 2 x − 1 = −1 x = 0 So, x = 2 & x = 0 Finding value of y If x = 2 y = 1/(𝑥 − 1) y = 1/(2 − 1) y = 1/1 y = 1 Thus, point is (2, 1) If x = 0 y = 1/(𝑥 − 1) y = 1/(0 − 1) y = 1/(−1) y = −1 Thus, point is (0, –1) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 1) and (0, −1) We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent through (2, 1) is 𝑦 −1 =−1 (𝑥 −2) 𝑦 −1 =−𝑥+2 𝒚+𝒙−𝟑 = 𝟎 Equation of tangent through (0, −1) is 𝑦 −(−1)=−1 (𝑥 −0) 𝑦 +1 =−𝑥 𝒚+𝒙 + 𝟏 = 𝟎