Tangents and Normals (using Differentiation)

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Chapter 6 Class 12 Application of Derivatives
Serial order wise

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

Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 7

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Transcript

Question 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 (𝐼𝑓 𝑦=0, (−16)/( 9) 𝑥/0=∞) 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.