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


Misc 24 The points on the curve 9y2 = π‘₯3, where the normal to the curve makes equal intercepts with the axes are (A) (4,Β±8/3) (B) (4,(βˆ’ 8)/3) (C) (4,Β±3/8) (D) (Β± 4, 3/8) Since Normal makes equal intercepts with the axes It’s equation will be π‘₯/π‘Ž+𝑦/𝑏=1 Putting b = a 𝒙/𝒂+π’š/𝒂=𝟏 π‘₯+𝑦=π‘Ž π’š=βˆ’π’™+𝒂 ∴ Slope of Normal = βˆ’1 Equation of line is π‘₯/π‘Ž+𝑦/𝑏=1 where a is x –intercept & b is y – intercept Now, finding slope of normal by Differentiation 9y2 = π‘₯3 Differentiating w.r.t π‘₯ 𝑑(9𝑦2)/𝑑π‘₯ = 𝑑(π‘₯3)/𝑑π‘₯ 9 𝑑(𝑦2)/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦=3π‘₯2 9 𝑑(𝑦2)/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ = 3x2 9(2𝑦) Γ— 𝑑𝑦/𝑑π‘₯ = 3x2 𝑑𝑦/𝑑π‘₯ = 3π‘₯2/9(2𝑦) π’…π’š/𝒅𝒙= π’™πŸ/πŸ”π’š We know that Slope of tangent Γ— slope of normal = –1 π‘₯2/6𝑦 Γ— Slope of normal = –1 Slope of normal = (βˆ’πŸ”π’š)/π’™πŸ Since Normal is at point (𝒉,π’Œ) Hence, Slope of normal at (β„Ž,π‘˜) = (βˆ’πŸ”π’Œ)/π’‰πŸ Now, Slope of Normal = βˆ’1 (βˆ’6π‘˜)/β„Ž2=βˆ’1 6k = h2 Also, Point (β„Ž,π‘˜) is on the curve 9y2 =π‘₯^3 So, (𝒉,π’Œ) will satisfy the equation of curve Putting π‘₯ = h & y = k in equation 9k2 = h3 Now our equations are 6k = h2 …(1) 9k2 = h3 …(2) From (3) 6k = h2 k = π’‰πŸ/πŸ” Putting value of k in (4) 9k2 = h3 9(β„Ž2/6)^2= h3 9(β„Ž4/36)=β„Ž3 β„Ž4/4 = h3 β„Ž4/β„Ž3 = 4 h = 4 Putting value of h = 4 in (4) 9k2 = h3 9k2 = (4)3 k2 = 64/9 k = Β± √(64/9) k = Β± πŸ–/πŸ‘ Hence required point is (h, k) = (4 , (Β±8)/3) Hence correct answer is A

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.