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

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt 2$. Its equation is

$\begin{array}{1 1}(A)\;x^2-y^2=32 \\(B)\;\frac{x^2}{4} -\frac{y^2}{9}=1 \\(C)\;2x^2-3y^2=7 \\(D)\; none\;of\;these \end{array}$

1 Answer

$2ae=16$
$ae=8\qquad e=\sqrt 2$
$a=4 \sqrt 2$
$a^2=32$
$b^2=a^2(e^2-1)$
$\qquad= 32 (2-1)$
$\qquad=32$
Hence the hyperbola is $x^2-y^2=32$
Hence A is the correct answer.
answered Apr 21, 2014 by meena.p
 

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