# 6. 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.       $\frac{x^2}{100} + \frac{y^2}{400} =1$

Given

The equation of the ellipse

$\frac{x^2}{100} + \frac{y^2}{400} =1\Rightarrow \frac{x^2}{10^2} + \frac{y^2}{20^2} =1$

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

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

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

We get

$a=20$ and $b=10$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{20^2-10^2}$

$c=\sqrt{300}=10\sqrt{3}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,10\sqrt{3})\:and\:(0,-10\sqrt{3})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,20)\:and\:(0,-20)$

The length of the major axis:

$2a=2(20)=40$

The length of minor axis:

$2b=2(10)=20$

The eccentricity :

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

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(10)^2}{20}=\frac{200}{20}=10$

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