Check sibling questions

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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.