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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

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Transcript

Ex 6.3, 25 Find the equation of the tangent to the curve √(3π‘₯βˆ’2) which is parallel to the line 4x βˆ’ 2y + 5 = 0 . Let (β„Ž , π‘˜) be the point on Curve from tangent to be taken We know that Equation of tangent is 𝑑𝑦/𝑑π‘₯ 𝑦=√(3π‘₯ βˆ’2) Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑑(3π‘₯ βˆ’2)^(1/2))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=3/(2√(3π‘₯ βˆ’2)) Slope of tangent at (β„Ž , π‘˜) is 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((β„Ž , π‘˜) )=3/(2√(3β„Ž βˆ’ 2)) Given tangent is parallel to the line 4π‘₯βˆ’2𝑦+5 So , Slope of tangent = Slope of 4π‘₯βˆ’2π‘₯+5 Now, Given line is 4π‘₯βˆ’2𝑦+5=0 βˆ’2𝑦=βˆ’4π‘₯βˆ’5 2𝑦=4π‘₯+5 𝑦=(4π‘₯ + 5)/2 𝑦=2π‘₯+5/2 The above Equation is of the form 𝑦=π‘šπ‘₯+𝑐 where m is Slope of line ∴ Slope of line is 2 Now, Slope of tangent at (β„Ž , π‘˜)= Slope of line 4π‘₯βˆ’3𝑦+5=0 3/(2√(3β„Ž βˆ’2))=2 3=2 Γ—2√(3β„Ž βˆ’2) 3=4√(3β„Ž βˆ’2) Squaring Both Sides (3)^2=(4√(3β„Ž βˆ’2))^2 9=(4)^2 (√(3β„Ž βˆ’2))^2 9=16(3β„Ž βˆ’2) 9/16=3β„Ž βˆ’2 3β„Ž βˆ’2=9/16 3β„Ž=9/16+2 3β„Ž=(9 + 32)/16 3β„Ž=41/16 β„Ž=41/(16 Γ— 3) β„Ž=41/48 Now, 𝑦=√(3π‘₯ βˆ’2) Since Point (β„Ž , π‘˜) is on the Curve Point (β„Ž , π‘˜) Satisfies the Equation of Curve Putting π‘₯=β„Ž , 𝑦=π‘˜ π‘˜=√(3β„Ž βˆ’2) Finding k when β„Ž=41/48 π‘˜=√(3 Γ—41/48βˆ’2) =√(41/16βˆ’2)=√((41 βˆ’32)/16)=√(9/16)=3/4 Hence the point is (h, k) = (41/48 , 3/4) We know that Equation of line at (π‘₯1 , 𝑦1)& having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (41/48 , 3/4) & having Slope 2 is (π‘¦βˆ’3/4)=2(π‘₯βˆ’41/48) (4𝑦 βˆ’ 3)/4=2((48π‘₯ βˆ’ 41)/48) (4𝑦 βˆ’ 3)/4=(48π‘₯ βˆ’ 41)/24 24(4𝑦 βˆ’ 3)/4=48π‘₯βˆ’41 6(4π‘¦βˆ’3)=48π‘₯βˆ’41 24π‘¦βˆ’18=48π‘₯βˆ’41 48π‘₯βˆ’41βˆ’24𝑦+18=0 48π‘₯βˆ’24π‘¦βˆ’23=0 48π‘₯βˆ’24𝑦=23 Hence Required Equation of tangent is πŸ’πŸ–π’™βˆ’πŸπŸ’π’š=πŸπŸ‘

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.