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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 : b = 3, c = 4, centre at the origin; foci on the x axis

$\begin {array} {1 1} (A)\;\large\frac{x^2}{25}+\large\frac{y^2}{16}=1 & \quad (B)\;\large\frac{x^2}{16}+\large\frac{y^2}{25}=1 \\ (C)\;\large\frac{x^2}{25}-\large\frac{y^2}{16}=1 & \quad (D)\;\large\frac{x^2}{16}-\large\frac{y^2}{25}=1 \end {array}$

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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 b = 3 and c = 4.
The centre is at the origin and foci is on the x - axis.
Since the foci is on the x - axis, the major axis is along the x - axis.
Hence the equation of the ellipse is of the form :
$ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
$c = 4, b = 3$
$ \therefore a^2=b^2+c^2$
$ = 9+16=25$
$ \therefore a = 5$
Hence the equation of the ellipse is
$ \large\frac{x^2}{25}$$+\large\frac{y^2}{16}$$=1$
answered Jul 18, 2014 by thanvigandhi_1
 

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