If a triangle is inscribed in a rectangular hyperbola, its orthocentre lies

inside the curve

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Outside the curve

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on the curve

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none of these

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Solution

The correct option is C
on the curve

Let rectangular hyperbola
$x y = c^{2}$ ..........(i)
Let three points on Eq. (i) are
$A (c t_{1}, \frac{c}{t_{1}}), B (c t_{2}, \frac{c}{t_{2}}), A (c t_{3}, \frac{c}{t_{3}})$
Let orthocentre is P(h,k) then slope of AP $\times$ slope of BC =-1
$\Rightarrow \frac{k - \frac{c}{t_{1}}}{h - c t_{1}} \times \frac{\frac{c}{t_{3}} - \frac{c}{t_{2}}}{c t_{3} - c t_{2}} = - 1 \Rightarrow \frac{k - \frac{c}{t_{1}}}{h - c t_{1}} \times - \frac{1}{t_{2} t_{3}} = - 1 \Rightarrow k - \frac{c}{t_{1}} = h t_{2} t_{3} - c t_{1} t_{2} t_{3} . . . . . . . . . . . . . . . . . . (i i) Similarly, B P ⊥ A C t h e n k - \frac{c}{t_{2}} = h t_{3} t_{1} - c t_{1} t_{2} t_{3} . . . . . . . . . . . . . . . . . (i i i)$
Subtracting Eq. (iii) from Eq. (ii), then we get $h = - \frac{c}{t_{1} t_{2} t_{3}}$
Substituting the value h in Eq. (ii) then $k = - c t_{1} t_{2} t_{3}$
$∴$ Orthocentre is $(- \frac{c}{t_{1} t_{2} t_{3}}, - c t_{1} t_{2} t_{3})$
which lies on $x y = c^{2}$

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