# 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

Ex 6.3,1

Ex 6.3,2

Ex 6.3,3

Ex 6.3,4

Ex 6.3,5 Important

Ex 6.3,6

Ex 6.3,7 Important

Ex 6.3,8

Ex 6.3,9

Ex 6.3,10

Ex 6.3,11

Ex 6.3,12 Important

Ex 6.3,13

Ex 6.3,14

Ex 6.3,15 Important

Ex 6.3,16

Ex 6.3,17

Ex 6.3,18 You are here

Ex 6.3,19

Ex 6.3,20

Ex 6.3,21

Ex 6.3,22

Ex 6.3,23

Ex 6.3,24

Ex 6.3,25

Ex 6.3,26 Important

Ex 6.3,27

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

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.