Find the value

Given:

$4 x^{2}+9 y^{2}=1$

$\frac{x^{2}}{\frac{1}{4}}+\frac{y^{2}}{\frac{1}{9}}=1$ …(i)

Since,

$\frac{1}{4}>\frac{1}{9}$

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

$a^{2}=\frac{1}{4}$ and $b^{2}=\frac{1}{9}$

$\Rightarrow \mathrm{a}=\sqrt{\frac{1}{4}}$ and $\mathrm{b}=\sqrt{\frac{1}{9}}$

$\Rightarrow \mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{1}{3}$

(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 $=2 \mathrm{a}$

$=2 \times \frac{1}{2}$

$=1$ unit

(ii) To find: Coordinates of the Vertices

Clearly, $a>b$

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

$=\left(\frac{1}{2}, 0\right)$ and $\left(-\frac{1}{2}, 0\right)$

(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 $c$

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

$=\frac{1}{4}-\frac{1}{9}$

$\mathrm{c}^{2}=\frac{9-4}{36}$

$\mathrm{C}^{2}=\frac{5}{36}$

$\mathrm{C}=\frac{\sqrt{5}}{6}$ …(I)

$\therefore$ Coordinates of foci $=\left(\pm \frac{\sqrt{5}}{6}, 0\right)$

(iv) To find: Eccentricity

We know that,

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

$\Rightarrow \mathrm{e}=\frac{\frac{\sqrt{5}}{6}}{\frac{1}{2}}=\frac{\sqrt{5}}{6} \times 2=\frac{\sqrt{5}}{3}[$ from $(\mathrm{l})]$

(v) To find: Length of the Latus Rectum

We know that,

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

$=\frac{2 \times\left(\frac{1}{3}\right)^{2}}{\frac{1}{2}}$

$=\frac{\frac{2}{9}}{\frac{1}{2}}$

$=\frac{2}{9} \times 2$