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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 : Vertices $(\pm 5, 0), foci (\pm 4, 0)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{9}+\large\frac{y^2}{25}=1 & \quad (B)\;\large\frac{x^2}{25}+\large\frac{y^2}{9}=1 \\ (C)\;\large\frac{x^2}{9}-\large\frac{y^2}{25}=1 & \quad (D)\;\large\frac{x^2}{25}-\large\frac{y^2}{9}=1 \end {array}$

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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 vertices and foci are $( \pm 5, 0)$ and $(\pm 4, 0)$ respectively.
It is clear that the major axis is along the x - axis.
a = 5 and c = 4
$ \therefore b^2=a^2-c^2$
$b^2=25-16$
$b = \sqrt 9$
$ = 3$
Hence the equation of the ellipse is $ \large\frac{x^2}{25}$$+\large\frac{y^2}{9}$$=1$

 

answered Jul 16, 2014 by thanvigandhi_1
edited Jul 16, 2014 by thanvigandhi_1
 

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