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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation for the ellipse that satisfies the given conditions : Ends of major axis $ (0, \pm \sqrt 5 )$, ends of minor axis $(\pm 1, 0)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{5}+\large\frac{y^2}{1}=1 & \quad (B)\; \large\frac{x^2}{1}+\large\frac{y^2}{5}=1 \\ (C)\;\large\frac{x^2}{1}-\large\frac{y^2}{5}=1 & \quad (D)\;\large\frac{x^2}{5}-\large\frac{y^2}{1}=1 \end {array}$

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Toolbox:
  • If the major axis is along x - axis then the equation of ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • If the major axis is along y - axis then the equation of ellipse is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
  • $c = \sqrt{a^2+b^2}$ , where $c$ is the foci of the ellipse.
Step 1 :
It is given that the ends of major axis is $(0, \pm \sqrt 5)$ and ends of minor axis is $( \pm 1, 0)$
$ \therefore $ It is clear that the major axis is along the y - axis.
Hence the equation should be of the form $ \large\frac{x^2}{b^2}$$+ \large\frac{Y^2}{a^2}$$=1$
hence $a^2= (\sqrt 5 )^2$ and $ b^2 = (1)^2$
$ \therefore $ Equation of the required ellipse is
$ \large\frac{x^2}{1}$$+ \large\frac{y^2}{5}$$=1$
answered Jul 18, 2014 by thanvigandhi_1
 

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