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Find the equation for the ellipse that satisfies the given conditions : Foci $(\pm 3, 0), a = 4$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{16}+\large\frac{y^2}{7}=1 & \quad (B)\;\large\frac{x^2}{7}+\large\frac{y^2}{16}=1 \\ (C)\;\large\frac{x^2}{16}-\large\frac{y^2}{7}=1 & \quad (D)\;\large\frac{x^2}{7}-\large\frac{y^2}{16}=1 \end {array}$

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  • 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
Foci is $( \pm 3, 0)$ and a = 4.
Since the foci lies on the x - axis, the major axis should be along x - axis.
Hence the equation of ellipse should be of the form
$ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
$c^2=a^2-b^2$ or $b^2=a^2-c^2$
Now substituting the values for c and a we get,
$ = 7$
Hence the equation of the ellipse is
$ \large\frac{x^2}{16}$$+\large\frac{y^2}{7}$$=1$
answered Jul 18, 2014 by thanvigandhi_1

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