Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12




Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.3, 18 For the curve ๐ฆ=4๐ฅ3 โ2๐ฅ5, find all the points at which the tangent passes through the origin. Let (โ , ๐) be the Required Point on the Curve at which tangent is to be taken Given Curve is ๐ฆ=4๐ฅ^3โ2๐ฅ^5 Since Point (โ , ๐) is on the Curve โ (โ , ๐) will satisfy the Equation of Curve Putting ๐ฅ=โ , ๐ฆ=๐ ๐=4โ^3โ2โ^5 We know that Slope of tangent to the Curve is ๐๐ฆ/๐๐ฅ ๐ฆ=4๐ฅ^3โ2๐ฅ^5 Differentiating w.r.t. ๐ฅ ๐๐ฆ/๐๐ฅ=๐(4๐ฅ^(3 )โ 2๐ฅ^5 )/๐๐ฅ ๐๐ฆ/๐๐ฅ=12๐ฅ^2โ10๐ฅ^4 Since tangent is taken from (โ , ๐) Slope of tangent at (โ, ๐) is ใ๐๐ฆ/๐๐ฅโใ_((โ, ๐) )=12โ^2โ10โ^4 Now, Equation of tangent at (โ , ๐) & having Slope 12โ^2โ10โ^4 is (๐ฆโ๐)=12โ^2โ10โ^4 (๐ฅโโ) Also, Given tangent passes through Origin โด (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 โ๐)=12โ^2โ10โ^4 (0โโ) โ๐=12โ^2โ10โ^4 (โโ) โ๐=โ12โ^3+10โ^5 Now, Equation of tangent at (โ , ๐) & having Slope 12โ^2โ10โ^4 is (๐ฆโ๐)=12โ^2โ10โ^4 (๐ฅโโ) Also, Given tangent passes through Origin โด (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 โ๐)=12โ^2โ10โ^4 (0โโ) โ๐=12โ^2โ10โ^4 (โโ) โ๐=โ12โ^3+10โ^5 We know that Equation of line at (๐ฅ1 , ๐ฆ1)& having Slope m is ๐ฆโ๐ฆ1=๐(๐ฅโ๐ฅ1) Now our Equations are ๐=4โ^3โ2โ^5 โ๐=โ12โ^3+10โ^5 Adding (1) and (2) ๐โ๐=4โ^3โ2โ^5+(โ12โ^3+10โ^5 ) 0=4โ^3โ12โ^3โ2โ^5+10โ^5 0=โ 8โ^3+8โ^5 โ 8โ^3+8โ^5=0 โ 8โ^3 (1โโ^2 )=0 โ 8โ^3=0 โ=0 1โโ^2=0 โ^2=1 โ=ยฑ1 When ๐=๐ ๐=4โ^3โ2โ^5 ๐=4(0)^3โ2(0)^5 ๐=0โ0 ๐=0 Point is (๐ , ๐) When ๐=๐ ๐=4โ^3โ2โ^5 ๐=4(1)^3โ2(1)^5 ๐=4โ2 ๐=2 Point is (๐ , ๐) When ๐=โ๐ ๐=4โ^3โ2โ^5 ๐=4(โ1)^3โ2(โ1)^5 ๐=โ4โ(โ2) ๐=โ2 Point is (โ๐ , โ๐) Hence the Required point on the Curve are (0 , 0), (1 , 2) & (โ1 , โ2)
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