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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

The equation of the ellipse having foci $(0,1),(0,-1)$ and minor axis of length $\sqrt 5$ is ____________

$\begin{array}{1 1}\large\frac{4x^2}{5}+\frac{4y^2}{1}\normalsize =1\\\large\frac{4x^2}{1}+\frac{4y^2}{5}\normalsize =1\\\large\frac{5x^2}{4}+\frac{y^2}{4}\normalsize =1\\\large\frac{x^2}{4}+\frac{5y^2}{4}\normalsize =1\end{array} $

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A)
Toolbox:
  • General equation of an ellipse whose axis is along the minor axis $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ where $b > a$
  • $a^2=b^2(1-e^2)$
Answer : $\large\frac{4x^2}{1}+\frac{4y^2}{5}\normalsize =1$
Given foci is $(0,\pm 1)$
Clearly the ellipse lies along the minor axis.
Length of the minor axis is $\sqrt 5$
(ie) $2b=\sqrt 5$
$\therefore b=\large\frac{\sqrt 5}{2}$
Given foci is $be=1$
$\Rightarrow \large\frac{\sqrt 5}{2}$$\times e=1$
$\therefore e=\large\frac{2}{\sqrt 5}$
Hence $a^2=b^2(1-e^2)$
$a^2=\large\frac{5}{4}$$(1-\large\frac{4}{5})$
$a^2=\large\frac{1}{4}$ and $b^2=\large\frac{5}{4}$
Hence equation of the ellipse is $\large\frac{x^2}{\Large\frac{1}{4}}+\frac{y^2}{\Large\frac{5}{4}}$$=1$
$\Rightarrow \large\frac{4x^2}{1}+\frac{4y^2}{5}$$=1$ is the required equation.
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