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Eccentricity

If the normal...

Question

If the normal at $(c t^{1}, {\frac{c}{t}}^{1})$ on the hyperbola $x y = c^{2}$ cuts the hyperbola again at $(c t^{2}, {\frac{c}{t}}^{2})$ , then $t^{1} t^{2} =$

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-2

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-1

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Solution

The correct option is C -1
Given hyperbola is xy $= c^{2} \Rightarrow \frac{d y}{d x} = - \frac{c^{2}}{x^{2}}$
$∴$ Slope of the normal at $P (c t_{1}, \frac{c}{t_{1}})$ is $t_{1}^{2}$
Equation of the normal at P is $t_{1}^{2} x - y = c t_{1}^{3} - \frac{c}{t_{1}}$
This normal passes through $(c t_{2}, \frac{c}{t_{2}})$ , we have $c t_{2} t_{1}^{2} - \frac{c}{t_{2}} = c t_{1}^{2} - \frac{c}{t_{1}} \Rightarrow t_{1}^{2} t_{2} = - 1$

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