Question

Find the area bounded by the ellipse $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$ and the ordinates $x=0$and $x=ae$, where $b^2=a^2\left(1-e^2\right)$ and $e<1$.

Answer 1

Hint: Simplify the given ellipse equation and integrate within the given ordinate limits to find the area.

An ellipse of the form $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$ will meet the X-axis at (a, 0) and the Y-axis at (0, b). Let these points be P (a,0) and Q (0, b). It is symmetrical about the axes.
The ordinates given are $x=0$and $x=ae$ which will be parallel to the Y-axis as shown in the figure.
The shaded area is the area bounded by the ellipse and the given ordinates.

Required area = Area of the shaded region
= $2\times$Area of QOCD
=$2\times {\int }_0^{ae}ydx$ …(1)
The given equation is $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$. Let us find the value of y from this equation and substitute in equation (1).
$\begin{array}{c}{\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1\\ \begin{array}{l}\begin{array}{c}{\displaystyle \frac{y^2}{b^2}}=1-{\displaystyle \frac{x^2}{a^2}}\\ {\displaystyle \frac{y^2}{b^2}}={\displaystyle \frac{a^2-x^2}{a^2}}\end{array}\\ y^2={\displaystyle \frac{b^2}{a^2}}\left(a^2-x^2\right)\\ y=\pm \sqrt{{\displaystyle \frac{b^2}{a^2}}\left(a^2-x^2\right)}\\ y=\pm {\displaystyle \frac{b}{a}}\sqrt{\left(a^2-x^2\right)}\end{array}\end{array}$
Since, the area in equation (1) which is the area of QOCD is in the 1st quadrant. Hence, the value of y will be positive.
Hence, $y={\displaystyle \frac{b}{a}}\sqrt{\left(a^2-x^2\right)}$ …(2)
Substituting (2) in (1),
Required area =$2\times {\int }_0^{ae}ydx$
$$\begin{array}{l}=2\underset{0}{\overset{ae}{\int }}{\displaystyle \frac{b}{a}}\sqrt{a^2-x^2}dx\\ ={\displaystyle \frac{2b}{a}}\underset{0}{\overset{ae}{\int }}\sqrt{a^2-x^2}dx\end{array}$$ (Since a and b are constants)
We know that, $${\int }^{\sqrt{a^2-x^2}}dx={\displaystyle \frac{x\sqrt{a^2-x^2}}{2}}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}({\displaystyle \frac{x}{a}})+c$$
Using this in the previous step, we get
Required area = $${\displaystyle \frac{2b}{a}}[{\displaystyle \frac{1}{2}}x\sqrt{a^2-x^2}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}{\displaystyle \frac{x}{a}}{]}_0^{ae}$$ $$\begin{array}{c}\begin{array}{l}={\displaystyle \frac{2b}{a}}[({\displaystyle \frac{ae}{2}}\sqrt{a^2-{(ae)}^2}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}{\displaystyle \frac{ae}{a}})-({\displaystyle \frac{0}{2}}\sqrt{a^2-0}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}({\displaystyle \frac{0}{a}}))]\\ ={\displaystyle \frac{2b}{a}}[{\displaystyle \frac{ae}{2}}\sqrt{a^2-a^2{e}^2}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}(e)-0-{\displaystyle \frac{a^2}{2}}{\sin }^{-1}(0)]\\ ={\displaystyle \frac{2b}{a}}[{\displaystyle \frac{ae}{2}}\cdot a\sqrt{1-e^2}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}e-0]\end{array}\\ \begin{array}{l}={\displaystyle \frac{2b}{a}}[{\displaystyle \frac{a^{2}e}{2}}\sqrt{1-e^2}+{\displaystyle \frac{a^2}{2}}{\sin }^{-1}e]\\ ={\displaystyle \frac{2b}{a}}({\displaystyle \frac{a^2}{2}})[e\sqrt{1-e^2}+{\sin }^{-1}e]\\ =ab[e\sqrt{1-e^2}+{\sin }^{-1}e]\end{array}\end{array}$$
Required Area bounded by the ellipse $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$ and the ordinates $x=0$and $x=ae$
$$=ab[e\sqrt{1-e^2}+{\sin }^{-1}e]$$

Note: The required area can also be found by integrating the entire shaded area QOABCD instead of finding $2\times$Area of QOCD. It would be a little lengthier and more unnecessary because the given ellipse is symmetrical about the origin.