The area of the quadrilateral formed by the f | vedclass

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(D) For the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. The eccentricity $e_1 = \sqrt{1+\frac{9}{16}} = \frac{5}{4}$. The

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$50$

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(D) For the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. The eccentricity $e_1 = \sqrt{1+\frac{9}{16}} = \frac{5}{4}$. The foci are $(\pm ae_1, 0) = (\pm 5, 0)$.For the conjugate hyperbola $\frac{y^2}{9}-\frac{x^2}{16}=1$ (or $\frac{x^2}{-16} - \frac{y^2}{-9} = 1$),the equation is $\frac{y^2}{9}-\frac{x^2}{16}=1$. Here $a^2=9$ and $b^2=16$. The eccentricity $e_2 = \sqrt{1+\frac{16}{9}} = \frac{5}{3}$. The foci are $(0, \pm be_2) = (0, \pm 5)$.The four foci are $(\pm 5, 0)$ and $(0, \pm 5)$.These points form a rhombus with diagonals of length $10$ and $10$.Area of the quadrilateral $= \frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 10 	imes 10 = 50$ square units.

The area of the quadrilateral formed by the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its conjugate hyperbola is (in square units):

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