The normal to the rectangular hyperbola $x y = - c^{2}$ at the point $^{'} t_{1}^{'}$ meets the curve again at the point $^{'} t_{2}^{'}$ . The value of $t_{1}^{3} \cdot t_{2}$ is

1

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c

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- c

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- 1

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Solution

The correct option is D $- 1$
The equation of the normal $t_{1}$ is $y - \frac{c}{t_{1}} = t_{1}^{2} (x - c t_{1})$
If this passes through $(c t_{2}, \frac{c}{t_{2}})$
$\frac{c}{t_{2}} - \frac{c}{t_{t}} = t_{1}^{2} (c t_{2} - c t_{1})$
$\Rightarrow - \frac{1}{t_{t} t_{2}} = t_{1}^{2} \Rightarrow 1 + t_{t}^{3} t_{2} = 0$
$\Rightarrow t_{1}^{3} t_{2} = - 1$
Hence, option 'D' is correct.

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