# Misc 24 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 24 The points on the curve 9y2 = 3, 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 , be the point on the curv 9y2 = 3, where the normal makes equal intercept with the axes. We know that slope of tangent on the curve is 9y2 = 3 Differentiating w.r.t 9 2 = 3 9 2 =3 2 9 2 = 3x2 9 2 = 3x2 = 3 2 9 2 = 2 6 We know that Slope of tangent slope of normal = 1 2 6 Slope of normal = 1 Slope of normal = 1 2 6 Slope of normal = 6 2 Since Normal is at point , Hence slope of normal at , = 6 2 Equation of normal to the curve makes equal intercept with axes + =1 + =1 + =1 + = = + There above equation is of the form y = m + c Where m is slope of line Slope of normal = 1 Slope of normal at , = 1 From (1) & (2) 6 2 = 1 6k = h2 Also, Point , is on the curve 9y2 = 3 , will satisfy the equation of curve Putting = h & y = k in equation 9k2 = h3 Now our equations are 6k = h2 (3) 9k2 = h3 (4) From (3) 6k = h2 k = 2 6 k = 8 9 Putting value of k in (4) 9k2 = h3 9 2 6 2 = h3 9 4 36 = 3 4 4 = h3 4 3 = 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

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.