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, 10 Find the equation of all lines having slope โ1 that are tangents to the curve ๐ฆ=1/(๐ฅ โ 1) , ๐ฅโ 1. Equation of Curve is ๐ฆ=1/(๐ฅ โ 1) Slope of tangent is ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=๐(1/(๐ฅ โ 1))/๐๐ฅ (๐๐ฆ )/๐๐ฅ =(๐ )/๐๐ฅ (๐ฅโ1)^(โ1) (๐๐ฆ )/๐๐ฅ =โ1ใ ร (๐ฅโ1)ใ^(โ1โ1) (๐๐ฆ )/๐๐ฅ =โ(๐ฅโ1)^(โ2) ๐๐ฆ/๐๐ฅ=(โ 1)/(๐ฅ โ 1)^2 Given that slope = โ1 Hence, ๐๐ฆ/๐๐ฅ = โ1 โด (โ 1)/(๐ฅ โ 1)^2 =โ1 1/(๐ฅ โ 1)^2 =1 1=(๐ฅ โ 1)^2 (๐ฅ โ 1)^2=1 ๐ฅ โ 1=ยฑ1 x โ 1 = 1 x = 2 x โ 1 = โ1 x = 0 So, x = 2 & x = 0 Finding value of y If x = 2 y = 1/(๐ฅ โ 1) y = 1/(2 โ 1) y = 1/1 y = 1 Thus, point is (2, 1) If x = 0 y = 1/(๐ฅ โ 1) y = 1/(0 โ 1) y = 1/(โ1) y = โ1 Thus, point is (0, โ1) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 1) and (0, โ1) We know that Equation of line at (๐ฅ1 , ๐ฆ1)& having Slope m is ๐ฆโ๐ฆ1=๐(๐ฅโ๐ฅ1) Equation of tangent through (2, 1) is ๐ฆ โ1 =โ1 (๐ฅ โ2) ๐ฆ โ1 =โ๐ฅ+2 ๐+๐โ๐ = ๐ Equation of tangent through (0, โ1) is ๐ฆ โ(โ1)=โ1 (๐ฅ โ0) ๐ฆ +1 =โ๐ฅ ๐+๐ + ๐ = ๐
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