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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: Ends of Major Axis $\;(0, \pm\sqrt 5)$, Ends of Minor Axis $\;(\pm 1,0)$

$\begin{array}{1 1}\frac{x^2}{1}+\frac{y^2}{5}=1 \\ \frac{x^2}{5}+\frac{y^2}{1}=1 \\ \frac{x^2}{16}+\frac{y^2}{5}=1 \\ \frac{x^2}{9}+\frac{y^2}{25}=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 $b$.
  • $c = \sqrt {a^2 - b^2}$
  • Given the major and minor axis, we can substitute in the above equation and arrive at the equation.
<< Enter Text >>
Given Ends of Major Axis $\;(0, \pm\sqrt 5)$, Ends of Minor Axis $\;(\pm 1,0) \rightarrow a = \sqrt 5, \; b = 1$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{1^2}$$+\large\frac{y^2}{\sqrt {5}^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{1}$$+\large\frac{y^2}{5}$$=1$
answered Apr 2, 2014 by balaji.thirumalai
edited Apr 2, 2014 by balaji.thirumalai
 

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