The distance of the focus of x^{2}-y^{2}=4 fr | vedclass

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(D) The given equation of the hyperbola is $x^{2}-y^{2}=4$. Dividing by $4$,we get $\frac{x^{2}}{4}-\frac{y^{2}}{4}=1$. Here,$a^{2}=4$ and $b^{2}=4$.

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$\sqrt{2}$

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(D) The given equation of the hyperbola is $x^{2}-y^{2}=4$. Dividing by $4$,we get $\frac{x^{2}}{4}-\frac{y^{2}}{4}=1$. Here,$a^{2}=4$ and $b^{2}=4$. For a hyperbola,$b^{2}=a^{2}(e^{2}-1)$,so $4=4(e^{2}-1)$,which implies $e^{2}-1=1$,so $e^{2}=2$ and $e=\sqrt{2}$. The coordinates of the foci are $(\pm ae, 0) = (\pm 2\sqrt{2}, 0)$. The equations of the directrices are $x = \pm \frac{a}{e} = \pm \frac{2}{\sqrt{2}} = \pm \sqrt{2}$. Consider the focus $(2\sqrt{2}, 0)$ and the nearer directrix $x = \sqrt{2}$. The distance between them is $|2\sqrt{2} - \sqrt{2}| = \sqrt{2}$.

The distance of the focus of $x^{2}-y^{2}=4$ from the directrix which is nearer to it,is

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