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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: Length of major axis is $26$ and Foci is $(\pm5,0)$

$\begin{array}{1 1}\large\frac{x^2}{169}+\large\frac{y^2}{144}=1 \\ \large\frac{x^2}{169}+\large\frac{y^2}{25}=1 \\ \large\frac{x^2}{144}+\large\frac{y^2}{169}=1 \\ \large\frac{x^2}{25}+\large\frac{y^2}{169}=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$
  • $c= \sqrt {a^2 - b^2}$
  • Given the length of major axis and foci, we know a and c, so we can calculate be and we can substitute in the above equation and arrive at the equation.
Given Length of major axis is $26$ and Foci is $(\pm5,0)$, we get $2a = 26 \rightarrow a = 13$ and $c = 5$.
$c^2 = a^2 + b^2 \rightarrow 25 = 169 + b^2 \rightarrow b = \sqrt{144} = 12$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{13^2}$$+\large\frac{y^2}{12^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{169}$$+\large\frac{y^2}{144}$$=1$
answered Apr 3, 2014 by balaji.thirumalai
 

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