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

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-1

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-c

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Solution

The correct option is B
-1

Normal at $t_{1}$ meets the curve agiain at $t_{2}$ .
So, the normal passes through $(c t_{1}, \frac{c}{t_{1}}) and (c t_{2}, \frac{c}{t_{2}})$
slope of normal = $\frac{\frac{c}{t_{2}} - \frac{c}{t_{1}}}{c t_{2} - c t_{1}} = - \frac{1}{t_{1} t_{2}}$
Also equation of normal at ' $t_{1}$ ' is
$x t_{1}^{3} - y t_{1} - c t_{1}^{4} + c = 0$
$∴$ Slope of normal = $t_{1}^{2} = - \frac{1}{t_{1} t_{2}}$
$∴ t_{1}^{3} t_{2} = - 1$

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The normal to the rectangular hyperbola xy=c2 at the point 't1' meets the curve again at the point 't2'. Then the value of t31t2is

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