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Find the area of the region bounded by the ellipse $\large\frac{x^{2}}{9}+\frac{y^{2}}{5}$$=1$ between the two latus rectums.

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  • Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
  • If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Required area $=4 \times area $ in the first quadrant
$A=\int \limits_0^{ae} \; y dx $
$\qquad=\bigg[(e= \sqrt {1-\large\frac{b^2}{a^2}})\bigg]$
$\qquad=4 \int \limits_0^2 \sqrt {1-\large\frac{x^2}{9}}$$dx$
$\qquad=4 \large\frac{\sqrt 5}{3} $$\int \limits _0^2 \sqrt {3^2 -x^2}dx$
$\qquad=4 \large\frac{\sqrt 5}{3} $$ \bigg[\large\frac{3^2} {2} $$\sin ^{-1} \large\frac{x}{3}+\frac{x}{2} $$\sqrt {3^2-x^2}\bigg]_0^2$
$\qquad=4 \large\frac{\sqrt 5}{3} $$ \bigg[\large\frac{9} {2} $$\sin ^{-1} \large\frac{2}{3}$$+\sqrt 5\bigg]$sq.units
answered Aug 16, 2013 by meena.p

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