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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 $ (\pm 6, 0), foci (\pm 4, 0)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{36}-\large\frac{y^2}{20}=1 & \quad (B)\;\large\frac{x^2}{20}-\large\frac{y^2}{36}=1 \\ (C)\;\large\frac{x^2}{36}+\large\frac{y^2}{20}=1 & \quad (D)\;\large\frac{x^2}{20}+\large\frac{y^2}{36}=1 \end {array}$

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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 :
The given vertices are $(\pm 6, 0)$ and foci $(\pm 4, 0)$
It is clear thta the vertices are on the x axis.
Hence the major axis is along x - axis.
$ \therefore $ The equation of the ellipse should be of the form
$ \large\frac{x^2}{a^2}$$ + \large\frac{y^2}{b^2}$$=1$
$ b^2=a^2-c^2$
$ \Rightarrow b^2 = 36-16$
$ \therefore b = \sqrt{20}$
Hence the equation of ellipse is
$ \large\frac{x^2}{36}$$+\large\frac{y^2}{20}$$=1$
answered Jul 18, 2014 by thanvigandhi_1
 

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