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

Transcript

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 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﷯ ⇒ ﷐ℎ,𝑘﷯ will satisfy the equation of curve Putting 𝑥 = h & y = k in equation ⇒ 9k2 = h3 Now our equations are 6k = h2 …(3) 9k2 = h3 …(4) From (3) 6k = h2 k = ﷐ℎ2﷮6﷯ k = ﷐± 8﷮9﷯ 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