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, 24 Find the equations of the tangent and normal to the hyperbola ๐ฅ^2/๐^2 โ ๐ฆ^2/๐^2 = 1 at the point (๐ฅ0 , ๐ฆ0) We know that Slope of tangent is ๐๐ฆ/๐๐ฅ Finding ๐ ๐/๐ ๐ ๐ฅ^2/๐^2 โ๐ฆ^2/๐^2 =1 โ๐ฆ^2/๐^2 =1โ๐ฅ^2/๐^2 ๐ฆ^2/๐^2 =๐ฅ^2/๐^2 โ1 Differentiating w.r.t.๐ฅ ๐(๐ฆ^2/๐^2 )/๐๐ฅ=๐(๐ฅ^2/๐^2 โ1)/๐๐ฅ 1/๐^2 ๐(๐ฆ^2 )/๐๐ฅ=๐/๐๐ฅ (๐ฅ^2/๐^2 )โ๐(1)/๐๐ฅ 1/๐^2 ร ๐(๐ฆ^2 )/๐๐ฅ ร ๐๐ฆ/๐๐ฆ=1/๐^2 ๐(๐ฅ^2 )/๐๐ฅโ0 1/๐^2 ๐(๐ฆ^2 )/๐๐ฆ ร ๐๐ฆ/๐๐ฅ=1/๐^2 . 2๐ฅ 1/๐^2 ร2๐ฆ ร ๐๐ฆ/๐๐ฅ=1/๐^2 2๐ฅ ๐๐ฆ/๐๐ฅ=2๐ฅ/๐^2 ร ๐^2/2๐ฆ ๐๐ฆ/๐๐ฅ=(๐^2 ๐ฅ)/(๐^2 ๐ฆ) Slope of tangent at (๐ฅ0 , ๐ฆ0) is ใ๐๐ฆ/๐๐ฅโใ_((๐ฅ0 , ๐ฆ0) )=(๐^2 ๐ฅ0)/(๐^2 ๐ฆ0) We know that Slope of tangent ร Slope of Normal =โ1 (๐^2 ๐ฅ0)/(๐^2 ๐ฆ0) ร Slope of Normal =โ1 Slope of Normal = (โใ ๐ใ^2 ๐ฆ0)/(๐^2 ๐ฅ0) We know that Equation of line at (๐ฅ1 , ๐ฆ1)& having Slope m is ๐ฆโ๐ฆ1=๐(๐ฅโ๐ฅ1) Equation of tangent at (๐ฅ0 , ๐ฆ0) & having Slope (๐^2 ๐ฅ0)/(๐^2 ๐ฆ0) is (๐ฆโ๐ฆ0)=(๐^2 ๐ฅ0)/(๐^2 ๐ฆ0) (๐ฅโ๐ฅ0) ๐^2 ๐ฆ0(๐ฆโ๐ฆ0)=๐^2 ๐ฅ0 (๐ฅโ ๐ฅ0) ๐^2 (๐ฆ0๐ฆโ๐ฆ0^2 )=๐^2 (๐ฅ0 ๐ฅโ๐ฅ0^2 ) (๐ฆ0 ๐ฆ โ ๐ฆ0^2)/๐^2 =((๐ฅ0 ๐ฅ โ ๐ฅ0^2 ))/๐^2 (๐ฆ0 ๐ฆ )/๐^2 โ(๐ฆ0^2)/๐^2 =(๐ฅ0 ๐ฅ )/๐^2 โ(๐ฅ0^2)/๐^2 Equation of Normal at (๐ฅ0 ,๐ฆ0) & having Slope(โใ ๐ใ^2 ๐ฆ0)/(๐^2 ๐ฅ0) is (๐ฆโ๐ฆ0)=(โใ ๐ใ^2 ๐ฆ0)/(๐^(2 ) ๐ฅ0) (๐ฅโ๐ฅ0) ((๐ฆ โ ๐ฆ0))/(ใ ๐ใ^(2 ) ๐ฆ0)=(โ 1)/(๐^(2 ) ๐ฅ0) (๐ฅโ๐ฅ0) (๐ฆ โ ๐ฆ0)/(ใ ๐ใ^(2 ) ๐ฆ0)=(โ (๐ฅ โ ๐ฅ0))/(๐^(2 ) ๐ฅ0) (๐ โ ๐๐)/(ใ ๐ใ^(๐ ) ๐๐)+(๐ โ๐๐)/(ใ ๐ใ^(๐ ) ๐๐)=๐ (๐ฆ0 ๐ฆ )/๐^2 โ(๐ฅ0 ๐ฅ )/๐^2 =โ(๐ฅ0^2)/๐^2 +(๐ฆ0^2)/๐^2 ((๐ฆ0 ๐ฆ )/๐^2 โ(๐ฅ0 ๐ฅ )/๐^2 )=โ((๐ฅ0^2)/๐^2 โ(๐ฆ0^2)/๐^2 ) ((๐ฆ0 ๐ฆ )/๐^2 โ(๐ฅ0 ๐ฅ )/๐^2 )=โ1 (๐๐ ๐ )/๐^๐ โ(๐๐ ๐ )/๐^๐ =๐ Since point (๐ฅ_0 ,๐ฆ_0 ) lie on the Curve โด It will satisfy the Equation of Curve โด (๐ฅ0^2)/๐^2 โ(๐ฆ0^2)/๐^2 =1
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