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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.3, 18 For the curve ๐‘ฆ=4๐‘ฅ3 โˆ’2๐‘ฅ5, find all the points at which the tangent passes through the origin. Let (โ„Ž , ๐‘˜) be the Required Point on the Curve at which tangent is to be taken Given Curve is ๐‘ฆ=4๐‘ฅ^3โˆ’2๐‘ฅ^5 Since Point (โ„Ž , ๐‘˜) is on the Curve โ‡’ (โ„Ž , ๐‘˜) will satisfy the Equation of Curve Putting ๐‘ฅ=โ„Ž , ๐‘ฆ=๐‘˜ ๐‘˜=4โ„Ž^3โˆ’2โ„Ž^5 We know that Slope of tangent to the Curve is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฆ=4๐‘ฅ^3โˆ’2๐‘ฅ^5 Differentiating w.r.t. ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(4๐‘ฅ^(3 )โˆ’ 2๐‘ฅ^5 )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=12๐‘ฅ^2โˆ’10๐‘ฅ^4 Since tangent is taken from (โ„Ž , ๐‘˜) Slope of tangent at (โ„Ž, ๐‘˜) is ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((โ„Ž, ๐‘˜) )=12โ„Ž^2โˆ’10โ„Ž^4 Now, Equation of tangent at (โ„Ž , ๐‘˜) & having Slope 12โ„Ž^2โˆ’10โ„Ž^4 is (๐‘ฆโˆ’๐‘˜)=12โ„Ž^2โˆ’10โ„Ž^4 (๐‘ฅโˆ’โ„Ž) Also, Given tangent passes through Origin โˆด (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 โˆ’๐‘˜)=12โ„Ž^2โˆ’10โ„Ž^4 (0โˆ’โ„Ž) โˆ’๐‘˜=12โ„Ž^2โˆ’10โ„Ž^4 (โˆ’โ„Ž) โˆ’๐‘˜=โˆ’12โ„Ž^3+10โ„Ž^5 Now, Equation of tangent at (โ„Ž , ๐‘˜) & having Slope 12โ„Ž^2โˆ’10โ„Ž^4 is (๐‘ฆโˆ’๐‘˜)=12โ„Ž^2โˆ’10โ„Ž^4 (๐‘ฅโˆ’โ„Ž) Also, Given tangent passes through Origin โˆด (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 โˆ’๐‘˜)=12โ„Ž^2โˆ’10โ„Ž^4 (0โˆ’โ„Ž) โˆ’๐‘˜=12โ„Ž^2โˆ’10โ„Ž^4 (โˆ’โ„Ž) โˆ’๐‘˜=โˆ’12โ„Ž^3+10โ„Ž^5 We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1)& having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Now our Equations are ๐‘˜=4โ„Ž^3โˆ’2โ„Ž^5 โˆ’๐‘˜=โˆ’12โ„Ž^3+10โ„Ž^5 Adding (1) and (2) ๐‘˜โˆ’๐‘˜=4โ„Ž^3โˆ’2โ„Ž^5+(โˆ’12โ„Ž^3+10โ„Ž^5 ) 0=4โ„Ž^3โˆ’12โ„Ž^3โˆ’2โ„Ž^5+10โ„Ž^5 0=โˆ’ 8โ„Ž^3+8โ„Ž^5 โˆ’ 8โ„Ž^3+8โ„Ž^5=0 โˆ’ 8โ„Ž^3 (1โˆ’โ„Ž^2 )=0 โˆ’ 8โ„Ž^3=0 โ„Ž=0 1โˆ’โ„Ž^2=0 โ„Ž^2=1 โ„Ž=ยฑ1 When ๐’‰=๐ŸŽ ๐‘˜=4โ„Ž^3โˆ’2โ„Ž^5 ๐‘˜=4(0)^3โˆ’2(0)^5 ๐‘˜=0โˆ’0 ๐‘˜=0 Point is (๐ŸŽ , ๐ŸŽ) When ๐’‰=๐Ÿ ๐‘˜=4โ„Ž^3โˆ’2โ„Ž^5 ๐‘˜=4(1)^3โˆ’2(1)^5 ๐‘˜=4โˆ’2 ๐‘˜=2 Point is (๐Ÿ , ๐Ÿ) When ๐’‰=โˆ’๐Ÿ ๐‘˜=4โ„Ž^3โˆ’2โ„Ž^5 ๐‘˜=4(โˆ’1)^3โˆ’2(โˆ’1)^5 ๐‘˜=โˆ’4โˆ’(โˆ’2) ๐‘˜=โˆ’2 Point is (โˆ’๐Ÿ , โˆ’๐Ÿ) Hence the Required point on the Curve are (0 , 0), (1 , 2) & (โˆ’1 , โˆ’2)

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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.