Check sibling questions

Ex 6.3

Ex 6.3, 1 Deleted for CBSE Board 2023 Exams

Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,4 Deleted for CBSE Board 2023 Exams

Ex 6.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,6 Deleted for CBSE Board 2023 Exams

Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,8 Deleted for CBSE Board 2023 Exams

Ex 6.3,9 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,10 Deleted for CBSE Board 2023 Exams

Ex 6.3,11 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,12 Deleted for CBSE Board 2023 Exams

Ex 6.3,13 Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iii) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iv) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (v) Deleted for CBSE Board 2023 Exams

Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,16 Deleted for CBSE Board 2023 Exams

Ex 6.3,17 Deleted for CBSE Board 2023 Exams

Ex 6.3,18 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,19 Deleted for CBSE Board 2023 Exams

Ex 6.3,20 Deleted for CBSE Board 2023 Exams

Ex 6.3,21 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,22 Deleted for CBSE Board 2023 Exams

Ex 6.3,23 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,24 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,25 Deleted for CBSE Board 2023 Exams You are here

Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 6.3, 25 - Find equation of tangent to root 3x-2 parallel

Ex 6.3,25 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,25 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,25 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,25 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.3,25 - Chapter 6 Class 12 Application of Derivatives - Part 6


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 𝑑𝑦/𝑑𝑥 𝑦=√(3𝑥 −2) Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=(𝑑(3𝑥 −2)^(1/2))/𝑑𝑥 𝑑𝑦/𝑑𝑥=3/(2√(3𝑥 −2)) Slope of tangent at (ℎ , 𝑘) is 〖𝑑𝑦/𝑑𝑥│〗_((ℎ , 𝑘) )=3/(2√(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 + 5)/2 𝑦=2𝑥+5/2 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 3/(2√(3ℎ −2))=2 3=2 ×2√(3ℎ −2) 3=4√(3ℎ −2) Squaring Both Sides (3)^2=(4√(3ℎ −2))^2 9=(4)^2 (√(3ℎ −2))^2 9=16(3ℎ −2) 9/16=3ℎ −2 3ℎ −2=9/16 3ℎ=9/16+2 3ℎ=(9 + 32)/16 3ℎ=41/16 ℎ=41/(16 × 3) ℎ=41/48 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=41/48 𝑘=√(3 ×41/48−2) =√(41/16−2)=√((41 −32)/16)=√(9/16)=3/4 Hence the point is (h, k) = (41/48 , 3/4) We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent at (41/48 , 3/4) & having Slope 2 is (𝑦−3/4)=2(𝑥−41/48) (4𝑦 − 3)/4=2((48𝑥 − 41)/48) (4𝑦 − 3)/4=(48𝑥 − 41)/24 24(4𝑦 − 3)/4=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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.