# Ex 6.3,18 - 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 6.3,18 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=4𝑥3−2𝑥5 Since Point ℎ , 𝑘 is on the Curve ⇒ ℎ , 𝑘 will satisfy the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=4ℎ3−2ℎ5 We know that Slope of tangent to the Curve is 𝑑𝑦𝑑𝑥 𝑦=4𝑥3−2𝑥5 Differentiating w.r.t. 𝑥 𝑑𝑦𝑑𝑥=𝑑4𝑥3 − 2𝑥5𝑑𝑥 𝑑𝑦𝑑𝑥=12𝑥2−10𝑥4 Since tangent is taken from ℎ , 𝑘 Slope of tangent at ℎ, 𝑘 is 𝑑𝑦𝑑𝑥│ℎ, 𝑘=12ℎ2−10ℎ4 Now, Equation of tangent at ℎ , 𝑘 & having Slope 12ℎ2−10ℎ4 is 𝑦−𝑘=12ℎ2−10ℎ4𝑥−ℎ Also, Given tangent passes through Origin ⇒ 0 , 0 will satisfies the Equation of tangent Putting x = 0, y = 0 in equation 0 −𝑘=12ℎ2−10ℎ40−ℎ −𝑘=12ℎ2−10ℎ4−ℎ −𝑘=−12ℎ3+10ℎ5 Now our Equations are 𝑘=4ℎ3−2ℎ5 …(1) −𝑘=−12ℎ3+10ℎ5 …(2) Adding (1) and (2) 𝑘−𝑘=4ℎ3−2ℎ5+−12ℎ3+10ℎ5 0=4ℎ3−12ℎ3−2ℎ5+10ℎ5 0=− 8ℎ3+8ℎ5 − 8ℎ3+8ℎ5=0 − 8ℎ31−ℎ2=0 Hence the Required point on the Curve are 0 , 0, 1 , 2 & −1 , −2

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Ex 6.3,18 You are here

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