Question 6 - Miscellaneous - Chapter 8 Class 12 Application of Integrals
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 1 (ii) Important
Misc 2 Important
Misc 3 Important
Misc 4 (MCQ)
Misc 5 (MCQ) Important
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams You are here
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 Important Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 (MCQ) Deleted for CBSE Board 2025 Exams
Question 14 (MCQ) Important Deleted for CBSE Board 2025 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
Question 6 Find the area of the smaller region bounded by the ellipse π₯^2/π^2 +π¦^2/π^2 =1 & π₯/π + π¦/π = 1 Letβs first draw the figure π^π/π^π +π^π/π^π =π is an which is a equation ellipse with x-axis as principal axis And, π/π + π/π = 1 is a line passing through A (a, 0) and B (0, b) Required Area Required Area = Area OACB β Area OAB Area OACB Area OACB = β«_0^πβγπ¦ ππ₯γ π¦ β Equation of ellipse π₯^2/π^2 +π¦^2/π^2 =1 π¦^2/π^2 =1βπ₯^2/π^2 π¦^2=π^2 [1βπ₯^2/π^2 ] π¦=Β±β(π^2 [1βπ₯^2/π^2 ] ) π¦=Β± πβ(1βπ₯^2/π^2 ) As OACB is in 1st quadrant, Value of π¦ will be positive β΄ π¦=πβ(1βπ₯^2/π^2 ) Now, Area OACB =β«_0^πβγπβ(1βπ₯^2/π^2 )γ ππ₯ =bβ«_0^πβγβ((π^2 β π₯^2)/π^2 ) ππ₯" " γ =π/π β«_0^πβγβ(π^2βπ₯^2 ) ππ₯" " γ =π/π [1/2 π₯ β(π^2βπ₯^2 )+π^2/2 sin^(β1)β‘γπ₯/πγ ]_0^π =π/π [1/2.πβ(π^2βπ^2 )+π^2/2 sin^(β1)β‘γπ/πγβ(1/2 0β(π^2β0^2 )+π^2/2 sin^(β1)β‘γ0/πγ )] =π/π [0+π^2/2.γπππγ^(βπ)β‘πβ0β0] =π/π [0+π^2/2.π /π ] =π/π Γ π^2/2 " Γ " π/2 =( πππ )/4 =π/π [0+π^2/2.π /π ] =π/π Γ π^2/2 " Γ " π/2 =( πππ )/4 Area OAB Area OAB =β«_0^πβγπ¦ ππ₯γ π¦ β Equation of line π₯/π+π¦/π=1 π¦/π=1βπ₯/π π¦=π[1βπ₯/π] Therefore, Area OAB =β«_0^πβπ[1βπ₯/π]ππ₯ = γπ[π₯βπ₯^2/2π]γ_0^π = π[πβπ^2/2πβ[0β0^2/2π]] = π[πβπ/2β0] = πΓπ/2 =ππ/2 β΄ Area Required = Area OACB β Area OAB =( π ππ )/4βππ/2 =ππ/2 [π/2β1] =ππ/2 [(π β 2)/2] =ππ/π [π βπ] square units