Finding equation of tangent/normal when point and curve is given

Chapter 6 Class 12 Application of Derivatives
Concept wise

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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) 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