From any point (h, k) four normals can be drawn to the hyperbola $x^{2} - y^{2} = 8$ and if the four feet of normals are $(x_{r}, y_{r}) r = 1, 2, 3, 4$ then $x_{1} x_{2} x_{3} x_{4} =$

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–16

256

– 64

answer is B.

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Detailed Solution

$G i v e n h y p e r b o l a x^{2} - y^{2} = 8 w h i c h i s e q i v a l e n t t o x y = 4 L e t (2 t_{i}, \frac{2}{t_{i}}) i = 1, 2, 3, 4 b e p o i n t s o n h y p e r b o l a E q u a t i o n o f n o r m a l a t (2 t_{i}, \frac{2}{t_{i}}) i s y - \frac{2}{t_{i}} = {t_{i}}^{2} (x - 2 t_{i}) \Rightarrow y t_{i} - 2 = {t^{3}}_{i} (x - 2 t_{i}) \Rightarrow 2 {t^{4}}_{i} - x {t^{3}}_{i} + y t_{i} - 2 = 0 \Rightarrow t_{1} t_{2} t_{3} t_{4} = - 1 N o w x_{1} x_{2} x_{3} x_{4} = - 16$