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Find the equation of an ellipse that satisfies the following conditions: Vertices $\;(\pm5,0)$, Foci $\;(\pm4,0)$

$\begin{array}{1 1}\large\frac{x^2}{25}+\large\frac{y^2}{9}=1 \\ \large\frac{x^2}{9}+\large\frac{y^2}{25}=1 \\ \large\frac{x^2}{25}+\large\frac{y^2}{16}=1 \\\large\frac{x^2}{16}+\large\frac{y^2}{9}=1\end{array} $

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  • 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 $a = 5$ and $c = 4$.
$c= \sqrt {a^2 - b^2} \rightarrow 16 = \sqrt{25-b^2} \rightarrow b = \sqrt 9 = 3$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{5^2}$$+\large\frac{y^2}{3^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{25}$$+\large\frac{y^2}{9}$$=1$
answered Apr 2, 2014 by balaji.thirumalai
edited Apr 2, 2014 by balaji.thirumalai

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