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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 minor axis $16$, foci $(0, \pm 6).$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{100}+\large\frac{y^2}{64}=1 & \quad (B)\;\large\frac{x^2}{64}+\large\frac{y^2}{100}=1 \\ (C)\;\large\frac{x^2}{100}-\large\frac{y^2}{64}=1 & \quad (D)\;\large\frac{x^2}{64}-\large\frac{y^2}{100}=1 \end {array}$

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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 :
Given length of the minor axis = 16.
Foci = $(0, \pm 6)$
Since the foci is on the y - axis, the major axis is along y - axis.
Hence the equation of the ellipse should be of the form.
$ \large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
Step 2 :
Length of the minor axis = 16.
(i.e) $2b=16$
$ \Rightarrow b = 8$
also $c = 6$
$a^2=b^2+c^2$
$ = 8^2+6^2$
$ = 64 + 36$
$ = 100$
Hence the equation of the ellipse is
$ \large\frac{x^2}{64}$$+\large\frac{y^2}{100}$$=1$
answered Jul 18, 2014 by thanvigandhi_1
 

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