


Last updated at April 14, 2021 by Teachoo
Transcript
Ex 6.3, 12 Find the equations of all lines having slope 0 which are tangent to the curve ๐ฆ = 1/(๐ฅ2 โ2๐ฅ + 3) Given Curve is ๐ฆ = 1/(๐ฅ2 โ2๐ฅ + 3) Slope of tangent is ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=๐(1/(๐ฅ2 โ 2๐ฅ + 3))/๐๐ฅ ๐๐ฆ/๐๐ฅ=(๐(๐ฅ2 โ 2๐ฅ + 3)^(โ1))/๐๐ฅ ๐๐ฆ/๐๐ฅ=โ1(๐ฅ2โ2๐ฅ+3)^(โ2) . ๐(๐ฅ^2โ 2๐ฅ + 3)/๐๐ฅ ๐๐ฆ/๐๐ฅ=โ(๐ฅ2โ2๐ฅ+3)^(โ2) (2๐ฅโ2) ๐๐ฆ/๐๐ฅ=โ2(๐ฅ2โ2๐ฅ+3)^(โ2) (๐ฅโ1) ๐๐ฆ/๐๐ฅ=(โ2(๐ฅ โ 1))/(๐ฅ2 โ 2๐ฅ + 3)^2 Hence Slope of tangent is (โ2(๐ฅ โ 1))/(๐ฅ2 โ 2๐ฅ + 3)^2 Given Slope of tangent is 0 โ ๐๐ฆ/๐๐ฅ=0 (โ2(๐ฅ โ 1))/(๐ฅ2 โ 2๐ฅ + 3)^2 =0 โ2(๐ฅ โ 1)=0 ร(๐ฅ2 โ 2๐ฅ + 3)^2 โ2(๐ฅ โ 1)=0 (๐ฅโ1)=0 ๐ฅ=1 Finding y when ๐ฅ=1 ๐ฆ=1/(๐ฅ2 โ 2๐ฅ + 3) ๐ฆ=1/((1)^2 โ 2(1) + 3) ๐ฆ=1/(1 โ 2 + 3) ๐ฆ=1/(2 ) Point is (๐ , ๐/๐) Thus , tangent passes through (1 , 1/2) Equation of tangent at (1 , 1/2) & having Slope zero is (๐ฆ โ1/2)=0(๐ฅโ1) ๐ฆ โ1/2=0 ๐ฆ=1/2 Hence Equation of tangent is ๐=๐/๐ We know that Equation of time passing through (๐ฅ , ๐ฆ) & having Slope m is (๐ฆ โ๐ฆ1)=๐(๐ฅ โ๐ฅ2)
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