The area of the quadrilateral formed by the f | vedclass

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(B) The given hyperbolas are conjugate to each other. The foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $(\pm ae_1, 0)$,where $e_1

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$2(a^2 + b^2)$

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(B) The given hyperbolas are conjugate to each other. The foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $(\pm ae_1, 0)$,where $e_1 = \sqrt{1 + \frac{b^2}{a^2}}$.The foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ are $(0, \pm be_2)$,where $e_2 = \sqrt{1 + \frac{a^2}{b^2}}$.The quadrilateral formed by these four foci is a rhombus with diagonals of lengths $2ae_1$ and $2be_2$.The area of the rhombus is given by $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes (2ae_1) 	imes (2be_2) = 2abe_1e_2$.We have $e_1 = \frac{\sqrt{a^2 + b^2}}{a}$ and $e_2 = \frac{\sqrt{a^2 + b^2}}{b}$.Substituting these values,we get Area $= 2ab 	imes \frac{\sqrt{a^2 + b^2}}{a} 	imes \frac{\sqrt{a^2 + b^2}}{b} = 2(a^2 + b^2)$.

The area of the quadrilateral formed by the foci of the hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ is

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