Let $F_{1}, F_{2}$ are the foci of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ and $F_{3}, F_{4}$ are the foci of its conjugate hyperbola. If $e_{H} a n d e_{C}$ are their eccentricities respectively then the statement which holds true is:

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Their auxiliary circles will have the same equation

$e_{H} > e_{C}$

Their equations of the asymptotes are different

Area of the quadrilateral formed by their foci is 50sq. units.

answer is C.

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Detailed Solution

$\begin{array}{l} q = 16 [e_{H}^{2} - 1] \Rightarrow e_{H} = \frac{5}{4} \\ F 1, F_{2} \equiv [\pm 5, 0], 16 = 9 [e_{C}^{2} - 1] \Rightarrow e_{C} = \frac{5}{3} \\ F 3, F 4 \equiv (0, \pm 5) \end{array}$
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Area of F₁ F₃ F₂ F₄= $\frac{1}{2} \times 10 \times 10 = 50$