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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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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}{25}$$+\large\frac{y^2}{100}$$=1$

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

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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}{4}$$+\large\frac{y^2}{25}$$=1$
Since $4 \lt 25$, he minorr 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 {4} = 2$ and $a = \sqrt {25} = 5$
$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{25-4} = \sqrt{21} $
1) The coordinates of the foci are $(0, \sqrt{21})$ and $(0,-\sqrt{21})$
2) The coordinates of the vertex are $(0,5)$ and $(0,-5)$
3) The Length of the major axis, $2a = 2\times5 = 10$
4) The Length of the minor axis, $2b = 2 \times 2 = 4$
5) The Eccentricity, $e = \large\frac{c}{a} $$ = \large\frac{\sqrt{21}}{5}$
6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$ = \large\frac{2 \times 2^2}{5} $$ = \large\frac{8}{5}$
answered Apr 1, 2014 by balaji.thirumalai
 

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