Get live Maths 1-on-1 Classs - Class 6 to 12
Miscellaneous
Misc 1 (ii) Important
Misc 2 Deleted for CBSE Board 2023 Exams
Misc 3
Misc 4 Important
Misc 5 Important
Misc 6 Important
Misc 7
Misc 8
Misc 9 Important You are here
Misc 10
Misc 11 Important
Misc 12 Important Deleted for CBSE Board 2023 Exams
Misc 13
Misc 14 Important
Misc 15 Deleted for CBSE Board 2023 Exams
Misc 16 (MCQ)
Misc 17 (MCQ) Important
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Miscellaneous
Last updated at March 16, 2023 by Teachoo
Misc 9 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