# Find the coordinates of the foci, the vertices, the length of major 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 $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$

Solution:

The given equation is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.

Here, the denominator of $\frac{x^{2}}{36}$ is greater than the denominator of $\frac{y^{2}}{16}$.

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

On comparing the given equation with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we obtain $a=6$ and $b=4$.

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{36-16}=\sqrt{20}=2 \sqrt{5}$

Therefore,

The coordinates of the foci are $(2 \sqrt{5}, 0)$ and $(-2 \sqrt{5}, 0)$.

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

Length of major axis = 2a = 12

Length of minor axis = 2b = 8

Eccentricity, $e=\frac{c}{a}=\frac{2 \sqrt{5}}{6}=\frac{\sqrt{5}}{3}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 16}{6}=\frac{16}{3}$

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