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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The eccentricity of the hyperbola whose latus rectum is $8$ and conjugate axis is equal to half the distance between the foci is

$\begin{array}{1 1}(A)\;\frac{4}{3}\\(B)\;\frac{4}{\sqrt 3} \\(C)\;\frac{2}{\sqrt 3} \\(D)\;\text{none of these} \end{array}$

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1 Answer

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We have $\large\frac{2b^2}{a}$$=8$
and $2b=\large\frac{1}{2}$$(2 ae)$
$\therefore \large\frac{2}{a}\bigg( \large\frac{ae}{2} \bigg)^2$$=8$
$ae^2=16$ -----(1)
Now $\large\frac{2b^2}{a}$$=8$
$b^2= 4a$
$a^2(e^2-1)=4a$
$ae^2-a=4$
Substitute the value of $ae^2=16$ in eq(2)
$16-a=4$
$-a=4-16$
$-a=-12$
$a=12$
$ae^2=16$
$e^2=\large\frac{16}{12}$
$e^2=\large\frac{4}{3}$
$e=\large\frac{2}{\sqrt 3}$
Hence C is the correct answer.
answered Apr 11, 2014 by meena.p
 

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