CBSE Class 12 Sample Paper for 2021 Boards
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CBSE Class 12 Sample Paper for 2021 Boards
Last updated at Dec. 16, 2024 by Teachoo
![Find the area of the ellipse x^2 + 9y^2 = 36 using integration [Video]](https://cdn.teachoo.com/bb9b8aa1-02ea-4f01-82b6-042682e39fce/slide38.jpg)
Note : This is similar to Ex 8.1, 4 of NCERT – Chapter 8 Class 12 Application of Integration
Check the answer here https:// www.teachoo.com /3328/730/Ex-8.1--4---Find-area-bounded-by-ellipse-x2-16---y2-9--1/category/Ex-8.1/
Question 34 (Choice 2) Find the area of the ellipse π₯2 + 9π¦2 = 36 using integration Equation of Ellipse is :- π₯2 + 9π¦2 = 36 π₯^2/36+(9π¦^2)/36=1 π₯^2/36+π¦^2/4=1 π₯^2/6^2 +π¦^2/2^2 =1 Since Ellipse is symmetric along x and y-axis Area of ellipse = Area of ABCD = 4 Γ [Area Of OBC] = 2 Γ β«_0^6βγπ¦.γ ππ₯ Finding y π₯2 + 9π¦2 = 36 9π¦2 = 36 β π₯2 π¦^2=1/9 (36βπ₯^2 ) Taking square root on both sides y = Β± β(1/9 (36βπ₯^2 ) ) y = Β± 1/3 β(36βπ₯^2 ) Since OBC is above x-axis y will be positive β΄ π=π/π β(ππβπ^π ) Area of ellipse = 4 Γ β«_π^πβγπ.γ π π = 4 Γ β«_0^6βγ 1/3 β(36βπ₯^2 )γ ππ₯ = 4/3 β«_0^6ββ((6)^2βπ₯^2 ) ππ₯ It is of form β(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Replacing a by 6 we get = 4/3 [π₯/2 β((6)^2βπ₯^2 )+(6)^2/2 sin^(β1)β‘γ π₯/6γ ]_0^6 = 4/3 [6/2 β((6)^2β(6)^2 )+18 γ sinγ^(β1)β‘(6/6)β0/2 β((6)^2β(0)^2 )β18sin^(β1) (0/6)] = 4/3 [18 sin^(β1) (1)] = 4/3 Γ 18 Γπ/2 = 12π square units
