Find the area of the smaller region bounded by the ellipse 9x2+4y2=1 and the line 3x+2y=1

Solution: First we need to find the are of eclipse=πab=π⋅3⋅2=6πarea of first quadrent=41⋅6πrequired area=41⋅6π−21⋅3⋅2=23π−3unit2

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Find the area of the smaller region bounded by the ellipse {{{

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Find the area of the smaller region bounded by the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$

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Daniel Discussed

Find the area of the smaller region bounded by the ellipse

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

and the line

\frac{x}{3} + \frac{y}{2} = 1

14 mins ago

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14 mins ago

Solution: First we need to find the are of eclipse
=

π a b = π \cdot 3 \cdot 2 = 6 π

area of first quadrent
=

\frac{1}{4} \cdot 6 π

required area
=

\frac{1}{4} \cdot 6 π - \frac{1}{2} \cdot 3 \cdot 2

\frac{3 π}{2} - 3 u n i t^{2}

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Question Text	Find the area of the smaller region bounded by the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$
Answer Type	Text solution:1
Upvotes	151