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Find the area of the smaller region boundary by the ellipse \( \large\frac{x^2}{9}+\large\frac{y^2}{4}=1\) and the straight line \(\large \frac{x}{3}+\frac{y}{2}=1 \).
cbse
class12
modelpaper-2012
sec-c
q25
modelpaper-2014
q27
math
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asked
Feb 2, 2013
by
thanvigandhi_1
retagged
Mar 20, 2014
by
balaji.thirumalai
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This is a specific case of the problem solved here:
http://clay6.com/qa/1132/
In the above solution, it is proved that the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$and the line $\frac{x}{a}+\frac{y}{b}=1$, is $\frac{ab}{4}(\pi-2)sq. units.$
In this case, given $a = 3$, $b = 2$, the area = $\large\frac{3 \times 2 }{4}$$(\pi - 2) $$\approx 1.71$ sq units.
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Mar 20, 2014
by
balaji.thirumalai
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Find the area of the smaller region bounded by the ellipse $ \large\frac{x^2}{9}+\frac{y^2}{4}$$=1$ and the straight line $\large \frac{x}{3}+\frac{y}{2}=1$.
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Find the area of the smaller region bounded by the ellipse $\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}$$=1$ and the straight line $\large\frac{x}{a}+\frac{y}{b}$$=1$.
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