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Example 16 - Find equation of all lines having slope 2, tangent

Example 16 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 16 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 16 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Example 16 Find the equation of all lines having slope 2 and being tangent Equation of curve is y + 2/(π‘₯ βˆ’ 3) = 0 Differentiating both sides w.r.t x (𝑑𝑦 )/𝑑π‘₯ + (𝑑 )/𝑑π‘₯ (2/(π‘₯ βˆ’ 3))=0 (𝑑𝑦 )/𝑑π‘₯ =βˆ’(𝑑 )/𝑑π‘₯ (2/(π‘₯ βˆ’ 3)) (𝑑𝑦 )/𝑑π‘₯ =βˆ’2 (𝑑 )/𝑑π‘₯ (π‘₯βˆ’3)^(βˆ’1) (𝑑𝑦 )/𝑑π‘₯ =βˆ’2γ€– Γ— βˆ’(π‘₯βˆ’3)γ€—^(βˆ’1βˆ’1) (𝑑𝑦 )/𝑑π‘₯ =2(π‘₯βˆ’3)^(βˆ’2) (π’…π’š )/𝒅𝒙 =𝟐/(𝒙 βˆ’ πŸ‘)^𝟐 Given that slope = 2 𝑑𝑦/𝑑π‘₯ = 2 2/(π‘₯ βˆ’ 3)^2 = 2 1/(π‘₯ βˆ’ 3)^2 = 1 (π‘₯βˆ’3)^2 = 1 π‘₯βˆ’3 = Β±1 x βˆ’ 3 = 1 x = 4 x βˆ’ 3 = βˆ’ 1 x = 2 x βˆ’ 3 = βˆ’ 1 x = 2 So, x = 4 & x = 2 Finding value of y If x = 2 y = (βˆ’2)/(π‘₯ βˆ’ 3) y = (βˆ’2)/(2 βˆ’ 3) 𝑦=2 Thus, point is (2, 2) If x = 4 y = (βˆ’2)/(π‘₯ βˆ’ 3) y = (βˆ’2)/(4 βˆ’ 3) 𝑦=βˆ’2 Thus, point is (4, –2) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 2) and (4, βˆ’ 2) We know that Equation of line at (π‘₯1 , 𝑦1)& having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent through (2, 2) is 𝑦 βˆ’ 2 = 2 (π‘₯ βˆ’2) 𝑦 βˆ’ 2 = 2π‘₯ βˆ’ 4 π’šβˆ’πŸπ’™+𝟐 = 𝟎 Equation of tangent through (4, βˆ’2) is 𝑦 βˆ’(βˆ’2) = 2 (π‘₯ βˆ’4) 𝑦 + 2 = 2π‘₯ βˆ’ 8 π’š βˆ’ πŸπ’™ + 𝟏𝟎 = 𝟎 to the curve y + 2/(π‘₯ βˆ’ 3) = 0

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.