# Ex 6.3, 14 (v) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by

Last updated at Aug. 19, 2021 by

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Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) π₯=cosβ‘π‘, π¦=sinβ‘π‘ ππ‘ π‘= π/4At π= π /π x = cos π/4 = 1/β2 y = sin π/4 = 1/β2 β΄ At π‘=π/4 , the point is (1/β2 " ," 1/β2) Finding ππ¦/ππ₯ ππ¦/ππ₯=(ππ¦/ππ‘)/(ππ₯/ππ‘) Now, Hence, ππ¦/ππ₯=(ππ¦/ππ‘)/(ππ₯/ππ‘) π=πππβ‘π Differentiating w.r.t.π‘ ππ₯/ππ‘=π(cosβ‘π‘ )/ππ‘ ππ₯/ππ‘=βsinβ‘π‘ π=πππβ‘π Differentiating w.r.t. π‘ ππ¦/ππ‘=π(sinβ‘π‘ )/ππ‘ ππ¦/ππ‘=cosβ‘π‘ ππ¦/ππ₯=cosβ‘π‘/(βsinβ‘π‘ ) ππ¦/ππ₯=βcotβ‘π‘ At π=π /π γππ¦/ππ₯βγ_(π‘=π/4)=βπππ‘(π/4) =β1 β΄ Slope of tangent at (1/β2 " ," 1/β2) is β 1 We know that Slope of tangent Γ Slope of Normal =β1 β1 Γ Slope of Normal =β1 Slope of Normal =(β1)/(β1) Slope of Normal =1 Hence Slope of tangent is β 1 & Slope of Normal is 1 Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1/β2 " ," 1/β2) & having Slope β1 is π¦ β1/β2 =βπ₯+ 1/β2 π₯+π¦ =1/β2+ 1/β2 π₯+π¦ =2/β2 π₯+π¦ =β2 π+πββπ=π Equation of Normal at (1/β2 " ," 1/β2) & having Slope 1 is (π¦β1/β2)=1(π₯β1/β2) π¦ β1/β2 =π₯β 1/β2 π¦ =π₯β1/β2+ 1/β2 π¦ =π₯ π =π

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Ex 6.3, 14 (v) You are here

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.