Miscellaneous

Chapter 6 Class 12 Application of Derivatives
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,(β 8)/3) (C) (4,Β±3/8) (D) (Β± 4, 3/8) Since Normal makes equal intercepts with the axes Itβs equation will be π₯/π+π¦/π=1 Putting b = a π/π+π/π=π π₯+π¦=π π=βπ+π β΄ Slope of Normal = β1 Equation of line is π₯/π+π¦/π=1 where a is x βintercept & b is y β intercept Now, finding slope of normal by Differentiation 9y2 = π₯3 Differentiating w.r.t π₯ π(9π¦2)/ππ₯ = π(π₯3)/ππ₯ 9 π(π¦2)/ππ₯ Γ ππ¦/ππ¦=3π₯2 9 π(π¦2)/ππ¦ Γ ππ¦/ππ₯ = 3x2 9(2π¦) Γ ππ¦/ππ₯ = 3x2 ππ¦/ππ₯ = 3π₯2/9(2π¦) ππ/ππ= ππ/ππ We know that Slope of tangent Γ slope of normal = β1 π₯2/6π¦ Γ Slope of normal = β1 Slope of normal = (βππ)/ππ Since Normal is at point (π,π) Hence, Slope of normal at (β,π) = (βππ)/ππ Now, Slope of Normal = β1 (β6π)/β2=β1 6k = h2 Also, Point (β,π) is on the curve 9y2 =π₯^3 So, (π,π) will satisfy the equation of curve Putting π₯ = h & y = k in equation 9k2 = h3 Now our equations are 6k = h2 β¦(1) 9k2 = h3 β¦(2) From (3) 6k = h2 k = ππ/π Putting value of k in (4) 9k2 = h3 9(β2/6)^2= h3 9(β4/36)=β3 β4/4 = h3 β4/β3 = 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± π/π Hence required point is (h, k) = (4 , (Β±8)/3) Hence correct answer is A

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.