Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Ex 6.3

Ex 6.3, 1
Deleted for CBSE Board 2023 Exams

Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,4 Deleted for CBSE Board 2023 Exams

Ex 6.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,6 Deleted for CBSE Board 2023 Exams

Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,8 Deleted for CBSE Board 2023 Exams

Ex 6.3,9 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,10 Deleted for CBSE Board 2023 Exams

Ex 6.3,11 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,12 Deleted for CBSE Board 2023 Exams

Ex 6.3,13 Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iii) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iv) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (v) Deleted for CBSE Board 2023 Exams You are here

Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,16 Deleted for CBSE Board 2023 Exams

Ex 6.3,17 Deleted for CBSE Board 2023 Exams

Ex 6.3,18 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,19 Deleted for CBSE Board 2023 Exams

Ex 6.3,20 Deleted for CBSE Board 2023 Exams

Ex 6.3,21 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,22 Deleted for CBSE Board 2023 Exams

Ex 6.3,23 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,24 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,25 Deleted for CBSE Board 2023 Exams

Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

Last updated at Aug. 19, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

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