Maharashtra State BoardHSC Commerce 12th Board Exam
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Solve the following : Find the area of the ellipse x216+y29 = 1. - Mathematics and Statistics

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Sum

Solve the following :

Find the area of the ellipse `x^2/(16) + y^2/(9)` = 1.

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Solution

By the symmetry of the ellipse, required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.

Given equation of the ellipse is `x^2/(16) + y^2/(9)` = 1

∴ `y^2/(9) = 1 - x^2/(16)`

∴ y2 = `9(1 - x^2/16)`

= `(9)/(16)(16 - x^2)`

∴ y = `± (3)/(4)sqrt(16 - x^2)`

∴ y = `(3)/(4)sqrt(16 - x^2)` ...[∵ In first quadrant, y > 0]

∴ Required area = 4(area of the region OPQO)

= `4int_0^4y*dx`

= `4int_0^4 (3)/(4)sqrt(16 - x^2)*dx`

= `3int_0^4 sqrt((4)^2 - x^2)*dx`

= `3[x/2 sqrt((4)^2 - x^2) + (4)^2/(2)sin^-1 (x/4)]_0^4`

= `3{[4/2 sqrt((4)^2 - (4)^2) + (4)^2/(2)sin^-1(4/4)] - [0/2 sqrt((4)^2 - (0)^2) + (4)^2/2sin^-1(0/4)]}`

= 3{[0 + 8sin–1 (1)] – [0 + 0]}

= `3(8 xx pi/2)`
= 12π sq. units.

Concept: Standard Forms of Ellipse
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 7 Applications of Definite Integration
Miscellaneous Exercise 7 | Q 4.4 | Page 158
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