$P Q$ and $R S$ are two perpendicular chords of the rectangular hyperbola $x y = c^{2}$ . If $C$ is the centre of the rectangular hyperbola, then the product of the slopes of $C P, C Q, C R$ and $C S$ is equal to

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Solution

The correct option is B $1$
Let $t_{1}, t_{2}, t_{3}, t_{4}$ be the parameter
of the points $P, Q, R, S$ respectively. Then, the coordinates of $P, Q, R, S$ are
$(c t_{1}, \frac{c}{t_{1}}), (c t_{2}, \frac{c}{t_{2}}), (c t_{3}, \frac{c}{t_{3}}), (c t_{4}, \frac{c}{t_{4}})$ respectively.
Now $P Q$ is perpendicular to $R S$
$\Rightarrow \frac{\frac{c}{t_{2}} - \frac{c}{t_{1}}}{{c t}_{2} - {c t}_{1}} \times \frac{\frac{c}{t_{4}} - \frac{c}{t_{3}}}{{c t}_{4} - {c t}_{3}} = - 1$ $\Rightarrow - \frac{1}{t_{1} t_{2}} \times - \frac{1}{t_{3} t_{4}} = - 1$
$\Rightarrow t_{1} t_{2} t_{3} t_{4} = - 1 \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
$∴$ Product of the slope of
$C P, C Q, C R$ and $C S$ .
$= \frac{1}{{t_{1}}^{2}} \times \frac{1}{{t_{2}}^{2}} \times \frac{1}{{t_{3}}^{2}} \times \frac{1}{{t_{4}}^{2}} = \frac{1}{{t_{1}}^{2} {t_{2}}^{2} {t_{3}}^{2} {t_{4}}^{2}} = 1 [u s i n g (i)]$
Hence, option 'B' is correct.

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PQ and RS are two perpendicular chords of the rectangular hyperbola xy=c2. If C is the centre of the rectangular hyperbola, then the product of the slopes of CP,CQ,CR and CS is equal to

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