Question

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

Answer 1

Equation of the latus rectum are $x=\pm ae$
$a^2=9,b^2=5$
Therefore, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$
$\Rightarrow ae=3.\frac{2}{3}=2$
Thus the equation of the L.R. are $x=\pm 2$.
The required area is bounded by the ellipse, $x=-2$ and $x=2$.
Since the curve is symmetrical about both axes the required area is $4$ times the area in the first quadrant. i.e., the area bounded by the curve $\frac{x^2}{9}+\frac{y^2}{5}=1$ (or) $y=\frac{\sqrt{5}}{3}\sqrt{9-x^2},x=0,x=2$ and $x$-axis.
Required area $=4{\int }_0^{2}ydx=4{\int }_0^2\frac{\sqrt{5}}{3}.\sqrt{9-x^2}dx$
$=\frac{4\sqrt{5}}{3}{[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x}{3}\right)]}_0^2$
$=\frac{4\sqrt{5}}{3}[\sqrt{5}+\frac{9}{2}{\sin }^{-1}\left(\frac{2}{3}\right)]$