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The co-ordinate of the vertices of the rectangular hyperbola $xy=16$ are

\[\begin{array}{1 1}(1)(4 ,4 )(-4 ,-4)&(2)(2 ,8 )(-2 , -8 )\\(3)(4 ,0 )(-4 , 0 )&(4)(8 , 0 )(-8 , 0)\end{array}\]

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The equation of the given rectangular hyperbola is $xy= 16$
$c^2=16$
$\large\frac{a^2}{2}$$=16$
$a^2=32=> a= \sqrt {32}$
$a=\sqrt {16 \times 2}=>a=4 \sqrt 2$
$\therefore $Vertices are $A \bigg( \large\frac{4 \sqrt 2}{\sqrt 2} ,\frac{4 \sqrt 2}{\sqrt 2}\bigg)$
$A' \bigg( \large\frac{-4 \sqrt 2}{\sqrt 2} ,\frac{-4 \sqrt 2}{\sqrt 2}\bigg)$
=>$(4 ,4 )(-4 ,-4)$
Hence 1 is the correct answer.
answered May 16, 2014 by meena.p
 

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