**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 $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$

**Solution:**

The given equation is $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$ or $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{10^{2}}=1$.

Here, the denominator of $\frac{y^{2}}{100}$ is greater than the denominator of $\frac{x^{2}}{25}$.

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

On comparing the given equation with $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we obtain $b=5$ and $a=10$.

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{100-25}=\sqrt{75}=5 \sqrt{3}$

Therefore,

The coordinates of the foci are $(0, \pm 5 \sqrt{3})$.

The coordinates of the vertices are $(0, \pm 10)$.

Length of major axis $=2 a=20$

Length of minor axis $=2 b=10$

Eccentricity, $e=\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 100}{20}=10$

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