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

Transcript

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) Let (β„Ž,π‘˜ ) be the point on the curv 9y2 = π‘₯3, where the normal makes equal intercept with the axes. We know that slope of tangent on the curve is 𝑑𝑦/𝑑π‘₯ 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) 𝑑𝑦/𝑑π‘₯= π‘₯2/6𝑦 We know that Slope of tangent Γ— slope of normal = –1 π‘₯2/6𝑦 Γ— Slope of normal = –1 Slope of normal = (βˆ’1)/(π‘₯^2/6𝑦) Slope of normal = (βˆ’6𝑦)/π‘₯2 Since Normal is at point (β„Ž,π‘˜) Hence slope of normal at (β„Ž,π‘˜) = (βˆ’6π‘˜)/β„Ž2 Equation of normal to the curve makes equal intercept with axes π‘₯/π‘Ž+𝑦/𝑏=1 π‘₯/π‘Ž+𝑦/π‘Ž=1 (π‘₯ + 𝑦)/π‘Ž=1 (Putting π‘₯ = k & y = k) …(1) Equation of line is π‘₯/π‘Ž+𝑦/𝑏=1 where a is x –intercept & b is y – intercept π‘₯ + 𝑦 = π‘Ž 𝑦 = –π‘₯ + π‘Ž There above equation is of the form y = mπ‘₯ + c Where m is slope of line β‡’ Slope of normal = –1 β‡’ Slope of normal at (β„Ž,π‘˜) = –1 From (1) & (2) (βˆ’6π‘˜)/β„Ž2=βˆ’1 6k = h2 Also, Point (β„Ž,π‘˜) is on the curve 9y2 =π‘₯^3 …(2) …(3) β‡’ (β„Ž,π‘˜) will satisfy the equation of curve Putting π‘₯ = h & y = k in equation β‡’ 9k2 = h3 Now our equations are 6k = h2 9k2 = h3 From (3) 6k = h2 k = β„Ž2/6 …(4) 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± 8/3 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 9 years. He provides courses for Maths and Science at Teachoo.