Question 13 Marks
Find the area enclosed by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Draw the figure.
Answer
View full question & answer→The ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ can also be written as follows :
$\frac{x^2}{(5)^2}+\frac{y^2}{(4)^2}=1$
In the figure given below, area of the region $\text{ABA}^{\prime} {B}^{\prime} A$ enclosed by the ellipse $=4$ $($Area of the region $ \text{AOBA}$ enclosed by the given curve, $x-$ axis, ordinates $x=0, x=5$ in the first quadrant$) ($since the ellipse is symmetrical about $x-$ axis and $y-$ axis both$)$
$=4 \int_0^5 y\ d x$
Here vertical strips have been taken.
Now we shall find the value of $y$ from the given equation of the ellipse.
$\frac{x^2}{(5)^2}+\frac{y^2}{(4)^2}=1 \Rightarrow \frac{y^2}{(4)^2}=1-\frac{x^2}{(5)^2}$
$\Rightarrow y^2=\frac{(4)^2}{(5)^2}\left((5)^2-x^2\right)$
$\Rightarrow y= \pm \frac{4}{5} \sqrt{(5)^2-x^2}$
Here we shall take only positive value of $y$ since the region $\text{AOBA}$ is in the first equadrant.
Hence required area
$=4 \int_0^5 \frac{4}{5} \sqrt{(5)^2-x^2} d x$
$=\frac{16}{5} \int_0^5 \sqrt{(5)^2-x^2} d x$
$=\frac{16}{5}\left(\frac{x}{5} \sqrt{(5)^2-x^2}+\frac{(5)^2}{2} \sin ^{-1} \frac{x}{5}\right)_0^5$
$\because \int \sqrt{a^2-x^2} d x =\frac{x}{a} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$
$ =\frac{16}{5}\left(0+\frac{(5)^2}{2} \sin ^{-1} 1-0-0\right)$
$ =\frac{16}{5}\left(\frac{25}{2} \times \frac{\pi}{2}\right)$
$ =20 \pi $ square units.
$\frac{x^2}{(5)^2}+\frac{y^2}{(4)^2}=1$
In the figure given below, area of the region $\text{ABA}^{\prime} {B}^{\prime} A$ enclosed by the ellipse $=4$ $($Area of the region $ \text{AOBA}$ enclosed by the given curve, $x-$ axis, ordinates $x=0, x=5$ in the first quadrant$) ($since the ellipse is symmetrical about $x-$ axis and $y-$ axis both$)$
$=4 \int_0^5 y\ d x$
Here vertical strips have been taken.
Now we shall find the value of $y$ from the given equation of the ellipse.
$\frac{x^2}{(5)^2}+\frac{y^2}{(4)^2}=1 \Rightarrow \frac{y^2}{(4)^2}=1-\frac{x^2}{(5)^2}$
$\Rightarrow y^2=\frac{(4)^2}{(5)^2}\left((5)^2-x^2\right)$
$\Rightarrow y= \pm \frac{4}{5} \sqrt{(5)^2-x^2}$
Here we shall take only positive value of $y$ since the region $\text{AOBA}$ is in the first equadrant.
Hence required area
$=4 \int_0^5 \frac{4}{5} \sqrt{(5)^2-x^2} d x$
$=\frac{16}{5} \int_0^5 \sqrt{(5)^2-x^2} d x$
$=\frac{16}{5}\left(\frac{x}{5} \sqrt{(5)^2-x^2}+\frac{(5)^2}{2} \sin ^{-1} \frac{x}{5}\right)_0^5$
$\because \int \sqrt{a^2-x^2} d x =\frac{x}{a} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$
$ =\frac{16}{5}\left(0+\frac{(5)^2}{2} \sin ^{-1} 1-0-0\right)$
$ =\frac{16}{5}\left(\frac{25}{2} \times \frac{\pi}{2}\right)$
$ =20 \pi $ square units.