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(D) Given that the vertices are $(\pm a, 0) = (\pm 3, 0)$,so $a = 3$. Given that the foci are $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$. Substituting $a=

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$x=3 \sec 	heta, y=\sqrt{7} 	an 	heta$

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(D) Given that the vertices are $(\pm a, 0) = (\pm 3, 0)$,so $a = 3$. Given that the foci are $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$. Substituting $a=3$,we get $3e = 4$,which implies $e = \frac{4}{3}$. We know the relation $b^2 = a^2(e^2 - 1)$. Substituting the values,$b^2 = 3^2 \left( (\frac{4}{3})^2 - 1 ight) = 9 \left( \frac{16}{9} - 1 ight) = 16 - 9 = 7$. Thus,$b = \sqrt{7}$. The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{9} - \frac{y^2}{7} = 1$. The parametric equations for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $x = a \sec 	heta$ and $y = b 	an 	heta$. Substituting $a=3$ and $b=\sqrt{7}$,we get $x = 3 \sec 	heta$ and $y = \sqrt{7} 	an 	heta$.

If the vertices and foci of a hyperbola are respectively $(\pm 3,0)$ and $(\pm 4,0)$,then the parametric equations of that hyperbola are:

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