Misc 9 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 12, 2019 by Teachoo
Last updated at Dec. 12, 2019 by Teachoo
Transcript
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
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