A hyperbola passes through the point P(sqrt{2 | vedclass

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(C) The equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Since the foci are $(\pm 2, 0)$,we have $ae = 2$,so $a^{2}e^{2} =

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$(2\sqrt{2}, 3\sqrt{3})$

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(C) The equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Since the foci are $(\pm 2, 0)$,we have $ae = 2$,so $a^{2}e^{2} = 4$.Using the relation $b^{2} = a^{2}(e^{2} - 1) = a^{2}e^{2} - a^{2}$,we get $b^{2} = 4 - a^{2}$,which implies $a^{2} + b^{2} = 4$.Since the hyperbola passes through $P(\sqrt{2}, \sqrt{3})$,we have $\frac{2}{a^{2}} - \frac{3}{b^{2}} = 1$.Substituting $a^{2} = 4 - b^{2}$,we get $\frac{2}{4 - b^{2}} - \frac{3}{b^{2}} = 1$.$2b^{2} - 3(4 - b^{2}) = b^{2}(4 - b^{2}) \Rightarrow 2b^{2} - 12 + 3b^{2} = 4b^{2} - b^{4}$.$b^{4} + b^{2} - 12 = 0 \Rightarrow (b^{2} + 4)(b^{2} - 3) = 0$.Since $b^{2} > 0$,we have $b^{2} = 3$,which gives $a^{2} = 4 - 3 = 1$.The equation of the hyperbola is $x^{2} - \frac{y^{2}}{3} = 1$.The tangent at $P(\sqrt{2}, \sqrt{3})$ is $\frac{\sqrt{2}x}{1} - \frac{\sqrt{3}y}{3} = 1$,which simplifies to $\sqrt{2}x - \frac{y}{\sqrt{3}} = 1$.Checking option $C$: $\sqrt{2}(2\sqrt{2}) - \frac{3\sqrt{3}}{\sqrt{3}} = 4 - 3 = 1$. Thus,it passes through $(2\sqrt{2}, 3\sqrt{3})$.

$A$ hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the tangent to this hyperbola at $P$ also passes through the point:

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