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Question

(b) $2ae = 16,$ $e = \sqrt 2 $

==> $a = 4\sqrt 2 $ and $b = 4\sqrt 2 $ 

equation is $\frac{{{x^2}}}{{{{(4\sqrt 2 )}^2}}} - \frac{{{y^2}}

Vedclass · Accepted Answer

${x^2} - {y^2} = 32$

Vedclass · Answer

(b) $2ae = 16,$ $e = \sqrt 2 $

==> $a = 4\sqrt 2 $ and $b = 4\sqrt 2 $ 

equation is $\frac{{{x^2}}}{{{{(4\sqrt 2 )}^2}}} - \frac{{{y^2}}}{{{{(4\sqrt 2 )}^2}}} = 1$

==> ${x^2} - {y^2} = 32$.

The equation of the hyperbola referred to its axes as axes of coordinate and whose distance between the foci is $16$ and eccentricity is $\sqrt 2 $, is

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