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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equation of the ellipse referred to its centre whose minor axis is equal to the distance between the foci and whose latus is 10?

$\begin{array}{1 1}(a)\;\large\frac{x^2}{100}+\frac{y^2}{50}\normalsize=1\\(b)\;\large\frac{x^2}{10}+\frac{y^2}{5}\normalsize=1\\(c)\;\large\frac{x^2}{200}+\frac{y^2}{100}\normalsize=1\\(d)\;\text{None of these}\end{array}$

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1 Answer

Let the equation of the ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1\qquad a > b$
Then the foci are $S(ae,0)$ and $S'(-ae,0)$
Length of minor axes =2b
Length of lactus rectum =$\large\frac{2b^2}{a}$
According to the question
$BB'=SS'$
$2b=2ae$
$b=ae$
$\large\frac{2b^2}{a}$$=10$
$b^2=5a$
$b^2=a^2(1-e^2)$
$a^2e^2=a^2(1-e^2)$
$e=\large\frac{1}{\sqrt 2}$
$b=\large\frac{a}{\sqrt 2}$
$b^2=\large\frac{a^2}{2}$
$5a=\large\frac{a^2}{2}$
$a=10$
$b^2=5\times 10=50$
Hence equation of ellipse
$\large\frac{x^2}{100}+\frac{y^2}{50}$$=1$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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