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Ex 6.3

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Ex 6.3, 19 - Find points on x2 + y2 - 2x - 3 = 0 tangents

Ex 6.3,19 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,19 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,19 - Chapter 6 Class 12 Application of Derivatives - Part 4


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‘₯𝑖𝑠Given 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 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 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.