If a triangle is inscribed in a rectangular hyperbola, the shortest distance of the orthocentre to the hyperbola is

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Solution

Let the parameters of the vertices $A, B$ and $C$ of the points on the hyperbola $x y = c^{2}$ be $t_{1}, t_{2}, t_{3}$ respectively.
Therefore, coordinates of $A, B$ and $C$ are $(c t_{1}, \frac{c}{t_{1}}), (c t_{2}, \frac{c}{t_{2}})$ and $(c t_{3}, \frac{c}{t_{3}})$ .

Slope of the line $B C$ is
$m = \frac{\frac{c}{t_{2}} - \frac{c}{t_{3}}}{c (t_{2} - t_{3})} \Rightarrow m = - \frac{1}{t_{2} t_{3}}$

Equation of the line passing through $A$ and perpedicular to $B C$ is
$y - \frac{c}{t_{1}} = t_{2} t_{3} (x - c t_{1}) \Rightarrow y - t_{2} t_{3} \cdot x = \frac{c}{t_{1}} - c t_{1} t_{2} t_{3} \dots (1)$
Similarly, any line through $B$ and perpedicular to $A C$ is
$y - t_{1} t_{3} \cdot x = \frac{c}{t_{2}} - c t_{1} t_{2} t_{3} \dots (2)$

Solving $(1)$ and $(2)$ ,
Coordinates of the orthocentre is $(\frac{- c}{t_{1} t_{2} t_{3}}, - c t_{1} t_{2} t_{3})$
This point lies on the hyperbola.
$∴$ Minimum distance is $0$

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