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Ex 6.3, 18 - For y = 4x3 - 2x5, find all points at which tangent

Ex 6.3,18 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,18 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,18 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,18 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Transcript

Ex 6.3, 18 For the curve 𝑦=4𝑥3 −2𝑥5, find all the points at which the tangent passes through the origin.Let (ℎ , 𝑘) be the Required Point on the Curve at which tangent is to be taken Given Curve is 𝑦=4𝑥^3−2𝑥^5 Since Point (ℎ , 𝑘) is on the Curve ⇒ (ℎ , 𝑘) will satisfy the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=4ℎ^3−2ℎ^5 We know that Slope of tangent to the Curve is 𝑑𝑦/𝑑𝑥 𝑦=4𝑥^3−2𝑥^5 Differentiating w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=𝑑(4𝑥^(3 )− 2𝑥^5 )/𝑑𝑥 𝑑𝑦/𝑑𝑥=12𝑥^2−10𝑥^4 Since tangent is taken from (ℎ , 𝑘) Slope of tangent at (ℎ, 𝑘) is 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=12ℎ^2−10ℎ^4 Now, Equation of tangent at (ℎ , 𝑘) & having Slope 12ℎ^2−10ℎ^4 is (𝑦−𝑘)=12ℎ^2−10ℎ^4 (𝑥−ℎ) Also, Given tangent passes through Origin ∴ (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 −𝑘)=12ℎ^2−10ℎ^4 (0−ℎ) −𝑘=12ℎ^2−10ℎ^4 (−ℎ) −𝑘=−12ℎ^3+10ℎ^5 We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Now our Equations are 𝑘=4ℎ^3−2ℎ^5 −𝑘=−12ℎ^3+10ℎ^5 Adding (1) and (2) 𝑘−𝑘=4ℎ^3−2ℎ^5+(−12ℎ^3+10ℎ^5 ) 0=4ℎ^3−12ℎ^3−2ℎ^5+10ℎ^5 0=− 8ℎ^3+8ℎ^5 − 8ℎ^3+8ℎ^5=0 − 8ℎ^3 (1−ℎ^2 )=0 − 8ℎ^3=0 ℎ=0 1−ℎ^2=0 ℎ^2=1 ℎ=±1 When 𝒉=𝟎 𝑘=4ℎ^3−2ℎ^5 𝑘=4(0)^3−2(0)^5 𝑘=0−0 𝑘=0 Point is (𝟎 , 𝟎) When 𝒉=𝟏 𝑘=4ℎ^3−2ℎ^5 𝑘=4(1)^3−2(1)^5 𝑘=4−2 𝑘=2 Point is (𝟏 , 𝟐) When 𝒉=−𝟏 𝑘=4ℎ^3−2ℎ^5 𝑘=4(−1)^3−2(−1)^5 𝑘=−4−(−2) 𝑘=−2 Point is (−𝟏 , −𝟐) Hence the Required point on the Curve are (0 , 0), (1 , 2) & (−1 , −2)

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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.