A triangle has its vertices on a rectangular hyperbola. Prove that the orthocentre of the triangle also lies on the same hyperbola.

Explanation

Consider hyperbola `xy=c^(2).`
Let the vertices of triangle ABC which are lying the hyperbola be
`A(ct_(1),(c)/(t_(1))),B(ct_(2),(c)/(t_(2)))and C(ct_(3),(c)/(t_(3)))`
To get the orthocentre, we find the equations of two altitudes and solve.
`"Slope of BC"=((c)/(t_(3))-(c)/(t_(2)))/(ct_(3)-ct_(2))=-(1)/(t_(2)t_(3))`
`therefore" Slope of altitude AD"=t_(2)t_(3)`

Equation of altitude AD is
`y-(c)/(t_(1))=t_(2)t_(3)(x-ct_(1))`
`"or "t_(1)y-c=t_(1)t_(2)t_(3)x-ct_(1)^(2)t_(2)t_(3)" "(1)`
Similarly, equation of altitude BE is
`t_(2)y-c=t_(1)t_(2)t_(3)x-ct_(1)t_(2)^(2)t_(3)" (2)"`
Solving (1) and (2), we get the orthocentre `H(-(c)/(t_(1)t_(2)t_(3)),-ct_(1)t_(2)t_(3)).`
Clearly, orthocentre lies on the same hyperbola, i.e., on `xy=c^(2).`