Question 14 (v) - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Tangents and Normals (using Differentiation)
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Tangents and Normals (using Differentiation)
Last updated at April 16, 2024 by Teachoo
Question 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