Find the value

Given:

$\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ …(i)

Since, $4<25$

So, above equation is of the form,

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

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

$a^{2}=25$ and $b^{2}=4$

$\Rightarrow a=\sqrt{25}$ and $b=\sqrt{4}$

$\Rightarrow a=5$ and $b=2$

(i) To find: Length of major axes

Clearly, $a

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

$=2 \times 5$

$=10$ units

(ii) To find: Coordinates of the Vertices

Clearly, $a>b$

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

$=(0,5)$ and $(0,-5)$

(iii) To find: Coordinates of the foci

We know that,

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

So, firstly we find the value of $c$

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

$=25-4$

$c^{2}=21$

$c=\sqrt{21} \ldots(i)$

$\therefore$ Coordinates of foci $=(0, \pm \sqrt{21})$

(iv) To find: Eccentricity

We know that,

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

$\Rightarrow \mathrm{e}=\frac{\sqrt{21}}{5}[$ from $(\mathrm{l})]$

(v) To find: Length of the Latus Rectum

We know that,

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

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

$=\frac{8}{5}$