# Find the area of the region bounded by the ellipse $\large\frac{x^{2}}{9}+\frac{y^{2}}{5}$$=1 between the two latus rectums. ## 1 Answer Toolbox: • Area bounded by the curve t=f(x), the x-axis and the ordinates x=a,x=b is \int \limits_a^b f(x) dx or \int \limits _a^b y dx • If the curve lies below the x-axis for a \leq x \leq b, then the area is \int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx Required area =4 \times area in the first quadrant A=\int \limits_0^{ae} \; y dx \qquad=\bigg[(e= \sqrt {1-\large\frac{b^2}{a^2}})\bigg] \qquad=4 \int \limits_0^2 \sqrt {1-\large\frac{x^2}{9}}$$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