दीर्घवृत्त में नाभियों और शीर्षों के निर्देशांक, दीर्घ और लघु अक्ष की लंबाइयाँ, उत्केंद्रता तथा नाभिलंब जीवा की लंबाई ज्ञात कीजिए

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

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 $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}$