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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.3, 19 Find the points on the curve π‘₯^2+𝑦^2 βˆ’2π‘₯ βˆ’3=0 at which the tangents are parallel to the π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 Given that Tangent is parallel to the π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 ∴ Slope of tangent = Slope of π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Finding π’…π’š/𝒅𝒙 π‘₯^2+𝑦^2 βˆ’2π‘₯ βˆ’3=0 Differentiating w.r.t.π‘₯ 𝑑(π‘₯^2 + 𝑦^2 βˆ’2π‘₯ βˆ’3)/𝑑π‘₯=0 𝑑(π‘₯^2 )/𝑑π‘₯+𝑑(𝑦^2 )/𝑑π‘₯βˆ’π‘‘(2π‘₯)/𝑑π‘₯βˆ’π‘‘(3)/𝑑π‘₯=0 2π‘₯+𝑑(𝑦^2 )/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯βˆ’2βˆ’0=0 𝑑(𝑦^2 )/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯=2βˆ’2π‘₯ 2𝑦 Γ— 𝑑𝑦/𝑑π‘₯=2βˆ’2π‘₯ 𝑑𝑦/𝑑π‘₯=(2 βˆ’ 2π‘₯)/2𝑦 𝑑𝑦/𝑑π‘₯=(2 (1 βˆ’ π‘₯))/2𝑦 𝑑𝑦/𝑑π‘₯=(1 βˆ’ π‘₯)/𝑦 Now, Since line is parallel to π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 Angle with π‘₯βˆ’π‘Žπ‘₯𝑖𝑠=0 πœƒ=0 Slope of π‘₯βˆ’π‘Žπ‘₯𝑖𝑠=tanβ‘πœƒ=tan⁑0Β°=0 Now Slope of tangent = Slope of π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 𝑑𝑦/𝑑π‘₯=0 (1 βˆ’ π‘₯)/𝑦=0 1βˆ’π‘₯=0 Γ— y 1βˆ’π‘₯=0 π‘₯=1 Finding y when π‘₯=1 π‘₯^2+𝑦^2βˆ’2π‘₯βˆ’3=0 (1)^2+𝑦^2βˆ’2(1)βˆ’3=0 1+𝑦^2βˆ’2βˆ’3=0 𝑦^2βˆ’4=0 𝑦^2=4 𝑦=±√4 𝑦=Β±2 Hence, the required points are (𝟏 , 𝟐) & (𝟏 , βˆ’πŸ)

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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.