**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 $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

**Solution:**

The given equation is $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ or $\frac{x^{2}}{2^{2}}+\frac{y^{2}}{5^{2}}=1$.

Here, the denominator of $\frac{y^{2}}{25}$ is greater than the denominator of $\frac{x^{2}}{4}$.

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 $b=2$ and $a=5$.

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{25-4}=\sqrt{21}$

Therefore,

The coordinates of the foci are $(0, \sqrt{21})$ and $(0,-\sqrt{21})$.

The coordinates of the vertices are (0, 5) and (0, –5)

Length of major axis = 2*a* = 10

Length of minor axis = 2*b* = 4

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{21}}{5}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}$

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