Check sibling questions

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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Ex 6.3, 18 For the curve 𝑦=4π‘₯3 βˆ’2π‘₯5, find all the points at which the tangent passes through the origin.Let (β„Ž , π‘˜) be the Required Point on the Curve at which tangent is to be taken Given Curve is 𝑦=4π‘₯^3βˆ’2π‘₯^5 Since Point (β„Ž , π‘˜) is on the Curve β‡’ (β„Ž , π‘˜) will satisfy the Equation of Curve Putting π‘₯=β„Ž , 𝑦=π‘˜ π‘˜=4β„Ž^3βˆ’2β„Ž^5 We know that Slope of tangent to the Curve is 𝑑𝑦/𝑑π‘₯ 𝑦=4π‘₯^3βˆ’2π‘₯^5 Differentiating w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(4π‘₯^(3 )βˆ’ 2π‘₯^5 )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=12π‘₯^2βˆ’10π‘₯^4 Since tangent is taken from (β„Ž , π‘˜) Slope of tangent at (β„Ž, π‘˜) is 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((β„Ž, π‘˜) )=12β„Ž^2βˆ’10β„Ž^4 Now, Equation of tangent at (β„Ž , π‘˜) & having Slope 12β„Ž^2βˆ’10β„Ž^4 is (π‘¦βˆ’π‘˜)=12β„Ž^2βˆ’10β„Ž^4 (π‘₯βˆ’β„Ž) Also, Given tangent passes through Origin ∴ (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 βˆ’π‘˜)=12β„Ž^2βˆ’10β„Ž^4 (0βˆ’β„Ž) βˆ’π‘˜=12β„Ž^2βˆ’10β„Ž^4 (βˆ’β„Ž) βˆ’π‘˜=βˆ’12β„Ž^3+10β„Ž^5 We know that Equation of line at (π‘₯1 , 𝑦1)& having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Now our Equations are π‘˜=4β„Ž^3βˆ’2β„Ž^5 βˆ’π‘˜=βˆ’12β„Ž^3+10β„Ž^5 Adding (1) and (2) π‘˜βˆ’π‘˜=4β„Ž^3βˆ’2β„Ž^5+(βˆ’12β„Ž^3+10β„Ž^5 ) 0=4β„Ž^3βˆ’12β„Ž^3βˆ’2β„Ž^5+10β„Ž^5 0=βˆ’ 8β„Ž^3+8β„Ž^5 βˆ’ 8β„Ž^3+8β„Ž^5=0 βˆ’ 8β„Ž^3 (1βˆ’β„Ž^2 )=0 βˆ’ 8β„Ž^3=0 β„Ž=0 1βˆ’β„Ž^2=0 β„Ž^2=1 β„Ž=Β±1 When 𝒉=𝟎 π‘˜=4β„Ž^3βˆ’2β„Ž^5 π‘˜=4(0)^3βˆ’2(0)^5 π‘˜=0βˆ’0 π‘˜=0 Point is (𝟎 , 𝟎) When 𝒉=𝟏 π‘˜=4β„Ž^3βˆ’2β„Ž^5 π‘˜=4(1)^3βˆ’2(1)^5 π‘˜=4βˆ’2 π‘˜=2 Point is (𝟏 , 𝟐) When 𝒉=βˆ’πŸ π‘˜=4β„Ž^3βˆ’2β„Ž^5 π‘˜=4(βˆ’1)^3βˆ’2(βˆ’1)^5 π‘˜=βˆ’4βˆ’(βˆ’2) π‘˜=βˆ’2 Point is (βˆ’πŸ , βˆ’πŸ) Hence the Required point on the Curve are (0 , 0), (1 , 2) & (βˆ’1 , βˆ’2)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

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.