# Misc 24 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

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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,(β 8)/3) (C) (4,Β±3/8) (D) (Β± 4, 3/8) Let (β,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 ππ¦/ππ₯ 9y2 = π₯3 Differentiating w.r.t π₯ π(9π¦2)/ππ₯ = π(π₯3)/ππ₯ 9 π(π¦2)/ππ₯ Γ ππ¦/ππ¦=3π₯2 9 π(π¦2)/ππ¦ Γ ππ¦/ππ₯ = 3x2 9(2π¦) Γ ππ¦/ππ₯ = 3x2 ππ¦/ππ₯ = 3π₯2/9(2π¦) ππ¦/ππ₯= π₯2/6π¦ We know that Slope of tangent Γ slope of normal = β1 π₯2/6π¦ Γ Slope of normal = β1 Slope of normal = (β1)/(π₯^2/6π¦) Slope of normal = (β6π¦)/π₯2 Since Normal is at point (β,π) Hence slope of normal at (β,π) = (β6π)/β2 Equation of normal to the curve makes equal intercept with axes π₯/π+π¦/π=1 π₯/π+π¦/π=1 (π₯ + π¦)/π=1 (Putting π₯ = k & y = k) β¦(1) Equation of line is π₯/π+π¦/π=1 where a is x βintercept & b is y β intercept π₯ + π¦ = π π¦ = βπ₯ + π There above equation is of the form y = mπ₯ + c Where m is slope of line β Slope of normal = β1 β Slope of normal at (β,π) = β1 From (1) & (2) (β6π)/β2=β1 6k = h2 Also, Point (β,π) is on the curve 9y2 =π₯^3 β¦(2) β¦(3) β (β,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 = β2/6 β¦(4) Putting value of k in (4) 9k2 = h3 9(β2/6)^2= h3 9(β4/36)=β3 β4/4 = h3 β4/β3 = 4 h = 4 Putting value of h = 4 in (4) 9k2 = h3 9k2 = (4)3 k2 = 64/9 k = Β± 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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Chapter 6 Class 12 Application of Derivatives

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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.