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Last updated at May 29, 2018 by Teachoo

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Ex 6.3,13 Find points on the curve 2 9 + 2 16 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. 2 9 + 2 16 = 1 2 16 =1 2 9 Differentiating w.r.t. 2 16 = 1 2 9 1 16 2 = 1 2 9 1 16 2 =0 1 9 2 1 16 2 = 1 9 2 1 16 2 = 1 9 2 = 1 9 2 1 16 2 = 16 9 Hence = 16 9 parallel to Given tangent is parallel to Slope of tangent = Slope of =0 16 9 =0 This is only possible if =0 when =0 2 4 + 2 16 =1 0 4 + 2 16 =1 2 16 =1 2 =16 = 4 Hence the points are , & , parallel to Similarly if line is parallel to Angle with =90 = 90 Slope = tan = tan 90 = Hence = 16 9 = 16 9 = 1 0 This will be possible only if Denominator is 0 9 =0 =0 Now it is given that 2 9 + 2 16 =1 Putting =0 2 9 + 0 16 =1 2 9 =1 2 =9 = 9 = 3 Hence the points at which tangent is parallel to are , & ,

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Ex 6.3,13 You are here

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.