Ex 6.3, 13 - Find points on x2/9 + y2/16 = 1 at which tangents - Ex 6.3

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.3,13 Find points on the curve ﷐﷐𝑥﷮2﷯﷮9﷯ + ﷐﷐𝑦﷮2﷯﷮16﷯ = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. ﷐﷐𝑥﷮2﷯﷮9﷯ + ﷐﷐𝑦﷮2﷯﷮16﷯ = 1 ﷐﷐𝑦﷮2﷯﷮16﷯=1−﷐﷐𝑥﷮2﷯﷮9﷯ Differentiating w.r.t. 𝑥 ﷐𝑑﷐﷐﷐𝑦﷮2﷯﷮16﷯﷯﷮𝑑𝑥﷯=﷐𝑑﷐1−﷐﷐𝑥﷮2﷯﷮9﷯﷯﷮𝑑𝑥﷯ ﷐1﷮16﷯﷐𝑑﷐﷐𝑦﷮2﷯﷯﷮𝑑𝑥﷯=﷐𝑑﷐1﷯﷮𝑑𝑥﷯−﷐𝑑﷐﷐﷐𝑥﷮2﷯﷮9﷯﷯﷮𝑑𝑥﷯ ﷐1﷮16﷯ × ﷐𝑑﷐﷐𝑦﷮2﷯﷯﷮𝑑𝑥﷯ × ﷐𝑑𝑦﷮𝑑𝑦﷯=0−﷐1﷮9﷯﷐𝑑﷐﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯ ﷐1﷮16﷯ × ﷐𝑑﷐﷐𝑦﷮2﷯﷯﷮𝑑𝑦﷯ × ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐− 1﷮9﷯﷐𝑑﷐﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯ ﷐1﷮16﷯ ×2𝑦 ×﷐𝑑𝑦﷮𝑑𝑥﷯=﷐− 1﷮ 9﷯2𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐﷐− 1﷮ 9﷯ 2𝑥﷮﷐1﷮16﷯ 2𝑦﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐− 16﷮9﷯ ﷐𝑥﷮𝑦﷯ Hence ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐− 16﷮9﷯ ﷐𝑥﷮𝑦﷯ parallel to 𝑥−𝑎𝑥𝑖𝑠 Given tangent is parallel to 𝑥−𝑎𝑥𝑖𝑠 ⇒ Slope of tangent = Slope of 𝑥−𝑎𝑥𝑖𝑠 ﷐𝑑𝑦﷮𝑑𝑥﷯=0 ﷐− 16﷮ 9﷯ ﷐𝑥﷮𝑦﷯=0 This is only possible if 𝑥=0 when 𝑥=0 ﷐﷐𝑥﷮2﷯﷮4﷯ + ﷐﷐𝑦﷮2﷯﷮16﷯=1 ﷐0﷮4﷯+﷐﷐𝑦﷮2﷯﷮16﷯=1 ﷐﷐𝑦﷮2﷯﷮16﷯=1 ﷐𝑦﷮2﷯=16 𝑦=±4 Hence the points are ﷐𝟎 , 𝟒﷯ & ﷐𝟎 , −𝟒﷯ parallel to 𝑦−𝑎𝑥𝑖𝑠 Similarly if line is parallel to 𝑦−𝑎𝑥𝑖𝑠 Angle with 𝑥−𝑎𝑥𝑖𝑠 =90° θ = 90° Slope = tan θ = tan 90°=𕔴 Hence ﷐𝑑𝑦﷮𝑑𝑥﷯=𕔴 ﷐16﷮9﷯ ﷐𝑥﷮𝑦﷯=𕔴 ﷐16𝑥﷮9𝑦﷯=﷐1﷮0﷯ This will be possible only if Denominator is 0 9𝑦=0 𝑦=0 Now it is given that ﷐﷐𝑥﷮2﷯﷮9﷯+﷐﷐𝑦﷮2﷯﷮16﷯=1 Putting 𝑦=0 ﷐﷐𝑥﷮2﷯﷮9﷯+﷐0﷮16﷯=1 ﷐﷐𝑥﷮2﷯﷮9﷯=1 ﷐𝑥﷮2﷯=9 𝑥=﷐﷮9﷯ 𝑥=±3 Hence the points at which tangent is parallel to 𝑦−𝑎𝑥𝑖𝑠 are ﷐𝟑 , 𝟎﷯ & ﷐−𝟑 , 𝟎﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.