Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


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


Ex 6.3, 13 Find points on the curve π‘₯^2/9 + 𝑦^2/16 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. π‘₯^2/9 + 𝑦^2/16 = 1 𝑦^2/16=1βˆ’π‘₯^2/9 Differentiating w.r.t. π‘₯ 𝑑(𝑦^2/16)/𝑑π‘₯=𝑑(1βˆ’ π‘₯^2/9)/𝑑π‘₯ 1/16 𝑑(𝑦^2 )/𝑑π‘₯=𝑑(1)/𝑑π‘₯βˆ’π‘‘(π‘₯^2/9)/𝑑π‘₯ 1/16 Γ— 𝑑(𝑦^2 )/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦=0βˆ’1/9 𝑑(π‘₯^2 )/𝑑π‘₯ 1/16 Γ— 𝑑(𝑦^2 )/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯=(βˆ’ 1)/9 𝑑(π‘₯^2 )/𝑑π‘₯ 1/16 Γ—2𝑦 ×𝑑𝑦/𝑑π‘₯=(βˆ’ 1)/( 9) 2π‘₯ 𝑑𝑦/𝑑π‘₯=((βˆ’ 1)/( 9) 2π‘₯)/(1/16 2𝑦) 𝑑𝑦/𝑑π‘₯=(βˆ’ 16)/9 π‘₯/𝑦 Hence 𝑑𝑦/𝑑π‘₯=(βˆ’ 16)/9 π‘₯/𝑦 parallel to π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 Given tangent is parallel to π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 β‡’ Slope of tangent = Slope of π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 𝑑𝑦/𝑑π‘₯=0 (βˆ’ 16)/( 9) π‘₯/𝑦=0 This is only possible if π‘₯=0 when π‘₯=0 π‘₯^2/4 + 𝑦^2/16=1 0/4+𝑦^2/16=1 (𝐼𝑓 𝑦=0, (βˆ’16)/( 9) π‘₯/0=∞) 𝑦^2/16=1 𝑦^2=16 𝑦=Β±4 Hence the points are (𝟎 , πŸ’) & (𝟎 , βˆ’πŸ’) parallel to π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 Similarly if line is parallel to π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 Angle with π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 =90Β° ΞΈ = 90Β° Slope = tan ΞΈ = tan 90Β°=∞ Hence 𝑑𝑦/𝑑π‘₯=∞ 16/9 π‘₯/𝑦=∞ 16π‘₯/9𝑦=1/0 This will be possible only if Denominator is 0 9𝑦=0 𝑦=0 Now it is given that π‘₯^2/9+𝑦^2/16=1 Putting 𝑦=0 π‘₯^2/9+0/16=1 π‘₯^2/9=1 π‘₯^2=9 π‘₯=√9 π‘₯=Β±3 Hence the points at which tangent is parallel to π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 are (πŸ‘ , 𝟎) & (βˆ’πŸ‘ , 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 9 years. He provides courses for Maths and Science at Teachoo.