The distance between the foci of a hyperbola | vedclass

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Question

(A) The distance between the foci of a hyperbola is given by $2ae = 16$. Given $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which simplifies to $a = \fr

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

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(A) The distance between the foci of a hyperbola is given by $2ae = 16$. Given $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which simplifies to $a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$. Thus,$a^{2} = (4\sqrt{2})^{2} = 32$. For a hyperbola,$b^{2} = a^{2}(e^{2} - 1)$. Substituting the values,$b^{2} = 32((\sqrt{2})^{2} - 1) = 32(2 - 1) = 32$. The standard equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$. Substituting $a^{2} = 32$ and $b^{2} = 32$,we get $\frac{x^{2}}{32} - \frac{y^{2}}{32} = 1$,which simplifies to $x^{2} - y^{2} = 32$.

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

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