The parameters of three points A,B,C on the rectangular hyperbola $xy = c^{2}$ are $t_{1}, t_{2}, t_{3} .$ If the area of $∆ ABC$ is equal to $\frac{k (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1})}{t_{1} t_{2} t_{3}}$ then 'k'

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$2 c^{2}$

$\frac{c^{2}}{4}$

$c^{2}$

$\frac{c^{2}}{2}$

answer is B.

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Detailed Solution

$G i v e n h y p e r b o l a x y = c^{2} L e t t_{1} (c t_{1}, \frac{c}{t_{1}}), t_{2} (c t_{2}, \frac{c}{t_{2}}) a n d t_{3} (c t_{3}, \frac{c}{t_{3}}) b e p o i n t s o n g i v e n h y p e r b o l a N o w a r e a o f ∆ A B C = a b s o l u t e v a l u e o f \frac{1}{2} |\begin{matrix} c t_{1} & \frac{c}{t_{1}} & 1 \\ c t_{2} & \frac{c}{t_{2}} & 1 \\ c t_{3} & \frac{c}{t_{3}} & 1 \end{matrix}| = \frac{1}{2} (c^{2}) |\begin{matrix} t_{1} & \frac{1}{t_{1}} & 1 \\ t_{2} & \frac{1}{t_{2}} & 1 \\ t_{3} & \frac{1}{t_{3}} & 1 \end{matrix}| R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} = \frac{c^{2}}{2} |\begin{matrix} t_{1} & \frac{1}{t_{1}} & 1 \\ t_{2} - t_{1} & \frac{t_{1} - t_{2}}{t_{1} t_{2}} & 0 \\ t_{3} - t_{1} & \frac{t_{1} - t_{3}}{t_{1} t_{3}} & 0 \end{matrix}| = \frac{c^{2}}{2} |\begin{matrix} t_{2} - t_{1} & \frac{t_{1} - t_{2}}{t_{1} t_{2}} \\ t_{3} - t_{1} & \frac{t_{1} - t_{3}}{t_{1} t_{3}} \end{matrix}| = \frac{c^{2}}{2} |t_{2} - t_{1} (\frac{t_{1} - t_{3}}{t_{1} t_{3}}) - (t_{3} - t_{1}) (\frac{t_{1} - t_{2}}{t_{1} t_{2}})| = \frac{c^{2}}{2} |\frac{(t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1})}{t_{1} t_{2} t_{3}}|$