Misc 24 - Chapter 6 Class 12 Application of Derivatives

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Misc 24 The points on the curve 9y2 = 𝑥3, where the normal to the curve makes equal intercepts with the axes are (A) (4,±8/3) (B) (4,(− 8)/3) (C) (4,±3/8) (D) (± 4, 3/8) Let (ℎ,𝑘 ) be the point on the curv 9y2 = 𝑥3, where the normal makes equal intercept with the axes. We know that slope of tangent on the curve is 𝑑𝑦/𝑑𝑥 9y2 = 𝑥3 Differentiating w.r.t 𝑥 𝑑(9𝑦2)/𝑑𝑥 = 𝑑(𝑥3)/𝑑𝑥 9 𝑑(𝑦2)/𝑑𝑥 × 𝑑𝑦/𝑑𝑦=3𝑥2 9 𝑑(𝑦2)/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 = 3x2 9(2𝑦) × 𝑑𝑦/𝑑𝑥 = 3x2 𝑑𝑦/𝑑𝑥 = 3𝑥2/9(2𝑦) 𝑑𝑦/𝑑𝑥= 𝑥2/6𝑦 We know that Slope of tangent × slope of normal = –1 𝑥2/6𝑦 × Slope of normal = –1 Slope of normal = (−1)/(𝑥^2/6𝑦) Slope of normal = (−6𝑦)/𝑥2 Since Normal is at point (ℎ,𝑘) Hence slope of normal at (ℎ,𝑘) = (−6𝑘)/ℎ2 Equation of normal to the curve makes equal intercept with axes 𝑥/𝑎+𝑦/𝑏=1 𝑥/𝑎+𝑦/𝑎=1 (𝑥 + 𝑦)/𝑎=1 (Putting 𝑥 = k & y = k) …(1) Equation of line is 𝑥/𝑎+𝑦/𝑏=1 where a is x –intercept & b is y – intercept 𝑥 + 𝑦 = 𝑎 𝑦 = –𝑥 + 𝑎 There above equation is of the form y = m𝑥 + c Where m is slope of line ⇒ Slope of normal = –1 ⇒ Slope of normal at (ℎ,𝑘) = –1 From (1) & (2) (−6𝑘)/ℎ2=−1 6k = h2 Also, Point (ℎ,𝑘) is on the curve 9y2 =𝑥^3 …(2) …(3) ⇒ (ℎ,𝑘) will satisfy the equation of curve Putting 𝑥 = h & y = k in equation ⇒ 9k2 = h3 Now our equations are 6k = h2 9k2 = h3 From (3) 6k = h2 k = ℎ2/6 …(4) Putting value of k in (4) 9k2 = h3 9(ℎ2/6)^2= h3 9(ℎ4/36)=ℎ3 ℎ4/4 = h3 ℎ4/ℎ3 = 4 h = 4 Putting value of h = 4 in (4) 9k2 = h3 9k2 = (4)3 k2 = 64/9 k = ± √(64/9) k = ± 8/3 Hence required point is (h, k) = (4 , (±8)/3) Hence correct answer is A