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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 : Length of major axis $26,$ foci $(\pm 5, 0)$

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

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Toolbox:
  • Length of the major axis is 2a
  • Length of the minor axis is 2b
  • Equation of an ellipse whose major axis is along x - axis is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • Equation of an ellipse whose minor axis is along y - axis is $ \large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
  • $c = \sqrt{a^2-b^2}$ whose $c$ is the foci.
Step 1 :
It is given that
Length of the major axis = 26.
Foci = $( \pm 5, 0)$
Since the foci is on the x - axis , the major axis is along x - axis.
Hence the equation of the ellipse should be of the form $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
length of the major axis = 26.
(i.e) 2a = 26
$ \Rightarrow a = 13 $
$ \therefore a^2 = 169$.
Step 2 :
$c^2=a^2-b^2$
$ \therefore b^2=a^2-c^2$
(i.e) $b^2=169-25$
$ \Rightarrow b = \sqrt{144}$
$ b = 12$
$ \therefore $ Equation of the ellipse is
$ \large\frac{x^2}{169}$$+\large\frac{y^2}{144}$$=1$

 

answered Jul 18, 2014 by thanvigandhi_1
 

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