Find the value

Given:

$x^{2}+4 y^{2}=100$

Divide by 100 to both the sides, we get

$\frac{1}{100} x^{2}+\frac{4}{100} y^{2}=1$

$\frac{x^{2}}{100}+\frac{y^{2}}{25}=1$ …(i)

Since, $100>25$

So, above equation is of the form,

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ …(ii)

Comparing eq. (i) and (ii), we get

$a^{2}=100$ and $b^{2}=25$

$\Rightarrow a=\sqrt{100}$ and $b=\sqrt{25}$

$\Rightarrow a=10$ and $b=5$

(i) To find: Length of major axes

Clearly, $a>b$, therefore the major axes of the ellipse is along $x$ axes.

$\therefore$ Length of major axes $=2 \mathrm{a}$

$=2 \times 10$

$=20$ units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

$\therefore$ Coordinate of vertices $=(a, 0)$ and $(-a, 0)$

$=(10,0)$ and $(-10,0)$

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci $=(\pm c, 0)$ where $c^{2}=a^{2}-b^{2}$

So, firstly we find the value of $c$

$c^{2}=a^{2}-b^{2}$

$=100-25$

$c^{2}=75$

$c=\sqrt{75}$

$c=5 \sqrt{3} \ldots(l)$

$\therefore$ Coordinates of foci $=(\pm 5 \sqrt{3}, 0)$

(iv) To find: Eccentricity

We know that,

Eccentricity $=\frac{\mathrm{c}}{\mathrm{a}}$

$\Rightarrow \mathrm{e}=\frac{5 \sqrt{3}}{10}=\frac{\sqrt{3}}{2}[$ from $(\mathrm{I})]$

(v) To find: Length of the Latus Rectum

We know that,

Length of Latus Rectum $=\frac{2 b^{2}}{a}$

$=\frac{2 \times(4)^{2}}{5}$

$=\frac{32}{5}$