# Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\large\frac{x^2}{100}$$+\large\frac{y^2}{400}$$=1$

$\begin{array}{1 1}Eccentricity =\frac{\sqrt 3}{2}\quad Latus\; Rectum = 10 \\ Eccentricity =\frac{\sqrt 3}{4}\quad Latus\; Rectum = 5 \\Eccentricity =\frac{\sqrt 3}{2}\quad Latus\; Rectum = \frac{5}{2} \\ Eccentricity =\frac{\sqrt 3}{4}\quad Latus \;Rectum =10 \end{array}$

Toolbox:
• Given an ellipse as follows:
• http://clay6.com/mpaimg/Toolbar_8.png
• The equation of the ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
• Compare the given equation to the general equation of the ellipse to infer $a$ and $b$.
• $C = \sqrt {a^2 - b^2}$
• Coordinates of foci are $(c,0)$ and $(-c,0)$
• The coordinates of the Vertex are $(a,0)$ and $(-a,0)$
• Length of major axis = $2a$, Length of minor axis = $2b$
• Eccentricity $e = \large\frac{c}{a}$
• Length of Latus Rectum = $\large\frac{2b^2}{a}$
Given: $\large\frac{x^2}{100}$$+\large\frac{y^2}{400}$$=1$
Since $4 \lt 25$, he minor axis is along the x-axis.
On comparing the equation with $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1 \rightarrow b = \sqrt {100} = 10$ and $a = \sqrt {400} = 20$
$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{400-100} = \sqrt{300} = 10\sqrt 3$
1) The coordinates of the foci are $(0, 10\sqrt{3})$ and $(0,-10\sqrt{3})$
2) The coordinates of the vertex are $(0,20)$ and $(0,-20)$
3) The Length of the major axis, $2a = 2\times20 = 40$
4) The Length of the minor axis, $2b = 2 \times 10 = 20$
5) The Eccentricity, $e = \large\frac{c}{a} $$= \large\frac{10\sqrt{3}}{20}$$=\large\frac{\sqrt 3}{2}$
6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$= \large\frac{2 \times 10^2}{20}$$ = 10$