Find the coordinates of the foci, the vertices, the length of major axis, the minor axis,

Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$

Solution:

The given equation is $36 x^{2}+4 y^{2}=144$.

It can be written as

$36 x^{2}+4 y^{2}=144$

Or, $\frac{x^{2}}{4}+\frac{y^{2}}{36}=1$

$\mathrm{Or}, \frac{x^{2}}{2^{2}}+\frac{y^{2}}{6^{2}}=1$ $\ldots(1)$

Here, the denominator of $\frac{y^{2}}{6^{2}}$ is greater than the denominator of $\frac{x^{2}}{2^{2}}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing equation (1) with $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we obtain $b=2$ and $a=6$.

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{36-4}=\sqrt{32}=4 \sqrt{2}$

Therefore,

The coordinates of the foci are $(0, \pm 4 \sqrt{2})$.

The coordinates of the vertices are (0, ±6).

Length of major axis = 2= 12

      Length of minor axis = 2b = 4

Eccentricity, $e=\frac{c}{a}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{6}=\frac{4}{3}$

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