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Last updated at Dec. 8, 2016 by Teachoo

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Ex 6.3,25 Find the equation of the tangent to the curve 3𝑥−2 which is parallel to the line 4x − 2y + 5 = 0 . Let ℎ , 𝑘 be the point on Curve from tangent to be taken We know that Equation of tangent is 𝑑𝑦𝑑𝑥 𝑦= 3𝑥 −2 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥= 𝑑 3𝑥 −2 12𝑑𝑥 𝑑𝑦𝑑𝑥= 32 3𝑥 −2 Slope of tangent at ℎ , 𝑘 is 𝑑𝑦𝑑𝑥│ ℎ , 𝑘= 32 3ℎ − 2 Given tangent is parallel to the line 4𝑥−2𝑦+5 So , Slope of tangent = Slope of 4𝑥−2𝑥+5 Now, Given line is 4𝑥−2𝑦+5=0 −2𝑦=−4𝑥−5 2𝑦=4𝑥+5 𝑦= 4𝑥 + 52 𝑦=2𝑥+ 52 The above Equation is of the form 𝑦=𝑚𝑥+𝑐 where m is Slope of line ∴ Slope of line is 2 Now, Slope of tangent at ℎ , 𝑘= Slope of line 4𝑥−3𝑦+5=0 32 3ℎ −2=2 3=2 ×2 3ℎ −2 3=4 3ℎ −2 Squaring Both Sides 32= 4 3ℎ −22 9= 42 3ℎ −22 9=16 3ℎ −2 916=3ℎ −2 3ℎ −2= 916 3ℎ= 916+2 3ℎ= 9 + 3216 3ℎ= 4116 ℎ= 4116 × 3 ℎ= 4148 Now, 𝑦= 3𝑥 −2 Since Point ℎ , 𝑘 is on the Curve Point ℎ , 𝑘 Satisfies the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘= 3ℎ −2 Finding k when ℎ= 4148 𝑘= 3 × 4148−2 = 4116−2= 41 −3216= 916= 34 Hence the point is (h, k) = 4148 , 34 Equation of tangent at 4148 , 34 & having Slope 2 is 𝑦− 34=2 𝑥− 4148 4𝑦 − 34=2 48𝑥 − 4148 4𝑦 − 34= 48𝑥 − 4124 24 4𝑦 − 34=48𝑥−41 6 4𝑦−3=48𝑥−41 24𝑦−18=48𝑥−41 48𝑥−41−24𝑦+18=0 48𝑥−24𝑦−23=0 48𝑥−24𝑦=23 Hence Required Equation of tangent is 𝟒𝟖𝒙−𝟐𝟒𝒚=𝟐𝟑

Ex 6.3

Ex 6.3, 1

Ex 6.3,2

Ex 6.3,3

Ex 6.3,4

Ex 6.3, 5 Important

Ex 6.3,6

Ex 6.3,7

Ex 6.3,8

Ex 6.3,9 Important

Ex 6.3,10

Ex 6.3,11 Important

Ex 6.3,12

Ex 6.3,13

Ex 6.3,14 Important

Ex 6.3,15 Important

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Ex 6.3,17

Ex 6.3,18 Important

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Ex 6.3,20

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Ex 6.3,22

Ex 6.3,23 Important

Ex 6.3,24 Important

Ex 6.3,25 You are here

Ex 6.3,26 Important

Ex 6.3,27

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 9 years. He provides courses for Maths and Science at Teachoo.