# 5.  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.       $\frac{x^2}{49} + \frac{y^2}{36} = 1$

Given

The equation of ellipse

$\frac{x^2}{49} + \frac{y^2}{36} = 1\Rightarrow \frac{x^2}{7^2} + \frac{y^2}{6^2} = 1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with standard equation of ellipse, which is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We get

$a=7$ and $b=6$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}$

$c=\sqrt{13}$

Hence,

Coordinates of the foci:

$(c,0)\:and\:(-c,0)=(\sqrt{13},0)\:and\:(-\sqrt{13},0)$

The vertices:

$(a,0)\:and\:(-a,0)=(7,0)\:and\:(-7,0)$

The length of major axis:

$2a=2(7)=14$

The length of minor axis:

$2b=2(6)=12$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{13}}{7}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(6)^2}{7}=\frac{72}{7}$

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