# Ex 6.3,25 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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

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Chapter 6 Class 12 Application of Derivatives

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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.