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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 $\;(\pm 13,0)$, Foci $\;(\pm 5,0)$

$\begin{array}{1 1}\frac{x^2}{144}+\frac{y^2}{169}=1 \\ \frac{x^2}{196}+\frac{y^2}{144}=1 \\ \frac{x^2}{144}+\frac{y^2}{225}=1 \\ \frac{x^2}{225}+\frac{y^2}{169}=1\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}{b^2}$$+\large\frac{y^2}{a^2}$$=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 $(-a,0)$
  • Given the Vertex and Foci, we can calculate $b$ and hence arrive at the equation of the ellipse.
Given Vertices and Foci, we get $a = 13$ and $c = 5$.
$c= \sqrt {a^2 - b^2} \rightarrow 25 = \sqrt{169-b^2} \rightarrow b = \sqrt 144 = 12$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{12^2}$$+\large\frac{y^2}{13^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{144}$$+\large\frac{y^2}{169}$$=1$
answered Apr 2, 2014 by balaji.thirumalai
 

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