From any point P(h, k), four normal can be drawn to the rectangular hyperbola $x y = c^{2}$ such that product of the abscissae of the feet of the normal = product of the ordinates of the feet of the normal is equal to

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Solution

The correct option is B
The equation of the normal to the hyperbola $x y = c^{2}$ at any point $(c t, \frac{c}{t}) i s x t^{3} - y t - c t^{4} + c = 0$
If it passes through P(h, k), then
$h t^{3} - k t - c t^{4} + c = 0$
$\Rightarrow c t^{4} - h t^{3} + k t - c = 0$ …… (i)
This being a fourth degree equation in t, gives four values of t, say $t_{1}, t_{2}, t_{3} a n d t_{4}$ such that corresponding to each value of t, there is a normal passing through P(h, k).
Let $A (c t_{1}, \frac{c}{t_{1}}), B (c t_{2}, \frac{c}{t_{2}}), C (c t_{3}, \frac{c}{t_{3}}) a n d D (c t_{4}, \frac{c}{t_{4}})$ be four points on the hyperbola such that normal at A, B, C and D passes through P.
Product of the abscissae of A, B, C and D
$= c t_{1} \times c t_{2} \times c t_{3} \times c t_{4} = c^{4} t_{1} t_{2} t_{3} t_{4}$
$= - c^{4}$ $[∵ t_{1} t_{2} t_{3} t_{4} = - 1]$
Product of the ordinates of A, B, C and D
$= \frac{c}{t_{1}} \times \frac{c}{t_{2}} \times \frac{c}{t_{3}} \times = \frac{c}{t_{4}}$
$= \frac{c^{4}}{t_{1} t_{2} t_{3} t_{4}} = - c^{4}$

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