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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The foci of a hyperbola coincide with the foci of the ellipse $\large\frac{x^2}{25}+\frac{y^2}{9}$$=1$.Find the equation of the hyperbola if its ecentricity is 3?

$\begin{array}{1 1}(a)\;\large\frac{x^2}{1}-\frac{y^2}{8}=\frac{16}{9}\\(b)\;\large\frac{x^2}{3}-\frac{y^2}{4}=\frac{15}{9}\\(c)\;x^2-y^2=9\\(d)\;\text{None of these}\end{array}$

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$\large\frac{x^2}{25}+\frac{y^2}{9}$$=1$
$e=\sqrt{1-\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{9}{25}}$
$\Rightarrow \large\frac{4}{5}$
Foci of ellipse are $(\pm ae,0)$ i.e $(\pm 4,0)$
Let 'e' be the ecentricity of the required hyperbola and its equation be
$\large\frac{x^2}{A^2}-\frac{y^2}{B^2}$$=1$
The co-ordinate of foci are $(\pm Ae',0)$
$(A\times e')=4$
$A\times 3=4$
$A=\large\frac{4}{3}$
$B^2=A^2(e'^2-1)$
$B^2=\large\frac{16\times 8}{9}$
Substituting the value of A & B we get the equation of hyperbola
$\large\frac{x^2}{1}-\frac{y^2}{8}=\frac{16}{9}$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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