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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 $ (0, \pm 13), foci (0, \pm 5)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{169}+\large\frac{y^2}{144}=1 & \quad (B)\;\large\frac{x^2}{144}+\large\frac{y^2}{169}=1 \\ (C)\;\large\frac{x^2}{169}-\large\frac{y^2}{144}=1 & \quad (D)\;\large\frac{x^2}{144}-\large\frac{y^2}{169}=1 \end {array}$

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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 :
Given coordinate for vertices and foci are $(0, \pm 13) $ and $(0, \pm 5)$ respectively.
It is clear that the vertices are on the y - axis.
Hence the major axis is along y - axis.
$ \therefore b^2=a^2-c^2$
(i.e) $b^2=16a-25$
$ \therefore b = \sqrt{144}$
$ = 12$
hence the equation of the ellipse is
$ \large\frac{x^2}{b^2}$$+ \large\frac{y^2}{a^2}$$=1$
(i.e) $ \large\frac{x^2}{144}$$+ \large\frac{y^2}{169}$$=1$
answered Jul 16, 2014 by thanvigandhi_1
 

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