Normals drawn at three different points $P, Q, R$ on rectangular hyperbola $x y = c^{2}$ intersect at some another point $S$ on the hyperbola. If centroid of the triangle $P Q R$ is $(a, b)$ , then $\frac{a + b}{c}$ equals :

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Solution

The correct option is A $0$
The equation of a normal at $(c t, \frac{c}{t})$ is given by $t y - t^{3} x + c t^{4} - c = 0$ .
This passes through another point $S$ $(c t^{'}, \frac{c}{t^{'}})$ on the hyperbola. Hence, we get $c t^{4} t^{'} - c t^{3} t^{' 2} + c t - c t^{'} = 0$
$\Rightarrow t_{1} + t_{2} + t_{3} + t^{'} = \frac{c t^{' 2}}{c t^{'}} = t^{'}$ (Sum of the roots)
So, $t_{1} + t_{2} + t_{3} = 0$ ....(1)
(where $t_{1}, t_{2}, t_{3}$ are parameters of points $P, Q$ and $R$ .)
Also $\sum t_{1} t_{2} = 0$
$\Rightarrow t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1} + t^{'} (t_{1} + t_{2} + t_{3}) = 0$
$\Rightarrow t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1} = 0$
Hence, the centroid of triangle $P Q R$ is the origin, and so $a + b = 0$ .

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