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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of an ellipse that satisfies the following conditions: Vertices $\;(\pm6,0)$, Foci $\;(\pm4,0)$

$\begin{array}{1 1} \frac{x^2}{36}+\frac{y^2}{20}=1 \\ \frac{x^2}{9}+\frac{y^2}{20}=1 \\ \frac{x^2}{36}+\frac{y^2}{16}=1 \\ \frac{x^2}{16}+\frac{y^2}{36}=1\end{array} $

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Toolbox:
  • Given an ellipse as follows:
  • http://clay6.com/mpaimg/Toolbar_7.png
  • The equation of the ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • $c= \sqrt {a^2 - b^2}$
  • Coordinates of foci are $(c,0)$ and $(-c,0)$
  • The coordinates of the Vertex are $(a,0)$ and $(-a,0)$
  • Given the Vertex and Foci, we can calculate $b$ and hence arrive at the equation of the ellipse.
Given Vertices and Foci, we get $a = 6$ and $c = 4$.
$c= \sqrt {a^2 - b^2} \rightarrow 16 = \sqrt{36-b^2} \rightarrow b = \sqrt {20} $
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{6^2}$$+\large\frac{y^2}{\sqrt{20}^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{36}$$+\large\frac{y^2}{20}$$=1$
answered Apr 2, 2014 by balaji.thirumalai
 

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