Find the area of the region bounded by the ellipse 4x2+9y2=1.

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Find the area of the region bounded by the ellipse {x^2}{4}+

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Find the area of the region bounded by the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ .

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Find the area of the region bounded by the ellipse

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1

15 mins ago

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15 mins ago

The given equation of the ellipse can be represented as

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1

the standard form of ellipse is given by

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

by comparing we get,

a = 2 an d b = 3

\Rightarrow y = 3 1 - \frac{x ^{2}}{4}

.............. (1)

It can be observed that the ellipse is symmetrical about

x

-axis and

y

-axis.

∴

Area bounded by ellipse

= 4 \times

Area

O A B

∴

Area of

O A B

= \int_{0}^{a} y d x

= \int_{0}^{2} y d x

= \int_{0}^{2} 3 1 - \frac{x ^{2}}{4} d x

[Using (1)]

= \frac{3}{2} \int_{0}^{2} 4 - x^{2} d x

= \frac{3}{2} [\frac{x}{2} 4 - x^{2} + \frac{4}{2} sin^{- 1} \frac{x}{2}]_{0}^{2}

= \frac{3}{2} [\frac{2 π}{2}] = \frac{3 π}{2}

Therefore, area bounded by the ellipse

= 4 \times \frac{3 π}{2} = 6 π

units

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Question Text	Find the area of the region bounded by the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ .
Answer Type	Text solution:1
Upvotes	150