# Ex 6.3,13 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3,13 Find points on the curve 𝑥29 + 𝑦216 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. 𝑥29 + 𝑦216 = 1 𝑦216=1−𝑥29 Differentiating w.r.t. 𝑥 𝑑𝑦216𝑑𝑥=𝑑1−𝑥29𝑑𝑥 116𝑑𝑦2𝑑𝑥=𝑑1𝑑𝑥−𝑑𝑥29𝑑𝑥 116 × 𝑑𝑦2𝑑𝑥 × 𝑑𝑦𝑑𝑦=0−19𝑑𝑥2𝑑𝑥 116 × 𝑑𝑦2𝑑𝑦 × 𝑑𝑦𝑑𝑥=− 19𝑑𝑥2𝑑𝑥 116 ×2𝑦 ×𝑑𝑦𝑑𝑥=− 1 92𝑥 𝑑𝑦𝑑𝑥=− 1 9 2𝑥116 2𝑦 𝑑𝑦𝑑𝑥=− 169 𝑥𝑦 Hence 𝑑𝑦𝑑𝑥=− 169 𝑥𝑦 parallel to 𝑥−𝑎𝑥𝑖𝑠 Given tangent is parallel to 𝑥−𝑎𝑥𝑖𝑠 ⇒ Slope of tangent = Slope of 𝑥−𝑎𝑥𝑖𝑠 𝑑𝑦𝑑𝑥=0 − 16 9 𝑥𝑦=0 This is only possible if 𝑥=0 when 𝑥=0 𝑥24 + 𝑦216=1 04+𝑦216=1 𝑦216=1 𝑦2=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 𝑑𝑦𝑑𝑥= 169 𝑥𝑦= 16𝑥9𝑦=10 This will be possible only if Denominator is 0 9𝑦=0 𝑦=0 Now it is given that 𝑥29+𝑦216=1 Putting 𝑦=0 𝑥29+016=1 𝑥29=1 𝑥2=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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Ex 6.3,13 You are here

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

Davneet Singh

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.