The normal the rectangle hyperbola xy= 4 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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Solution

$x y = 4$
$x \frac{d y}{d x} + y = 0$
$\Rightarrow \frac{d y}{d x} = - \frac{y}{x}$
Slope of normal at $A (t_{1})$ is $\frac{- d x}{d y} | = \frac{x}{y} = \frac{2 t_{1}}{2 / t_{1}}$
$(2 t_{1}, \frac{2}{t_{1}}) = t_{1}^{2}$
Equation of normal
$t_{1}^{2} (x - 2 t_{1}) = (y - \frac{2}{t_{1}})$
It passes through $B (t_{2}) \equiv (2 t_{2}, \frac{2}{t_{2}})$
$t_{1}^{2} (2 t_{2} - 2 t_{1}) = (\frac{2}{t_{2}} - \frac{2}{t_{1}})$
$t_{1}^{2} (t_{2} - t_{1}) = \frac{t_{1} - t_{2}}{t_{1} t_{2}}$ $(t_{1} \neq t_{2})$ As A & B are different points.
$t_{1}^{2} = \frac{- 1}{t_{1} t_{2}}$
$t_{1}^{2} t_{2} = - 1$

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