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.

Answer 1

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 $

http://clay6.com/mpaimg/tn%207.4%207.JPG

$\qquad=\bigg[(e= \sqrt {1-\large\frac{b^2}{a^2}})\bigg]$

$\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