Question
If the coordinates of four concyclic points on the rectangular hyperbola $$xy = {c^2}$$ are $$\left( {c{t_{\text{i}}},\,\frac{c}{{{t_{\text{i}}}}}} \right),\,{\text{i}} = 1,\,2,\,3,\,4$$ then :
A.
$${t_1}{t_2}{t_3}{t_4} = - 1$$
B.
$${t_1}{t_2}{t_3}{t_4} = 1$$
C.
$${t_1}{t_3} = {t_2}{t_4}$$
D.
$${t_1} + {t_2} + {t_3} + {t_4} = {c^2}$$
Answer :
$${t_1}{t_2}{t_3}{t_4} = 1$$
Solution :
Let the points lie on the circle $${x^2} + {y^2} + 2gx + 2fy + 2fy + k = 0,$$ then
$$\eqalign{
& {c^2}t_{\text{i}}^2 + \frac{{{c^2}}}{{t_{\text{i}}^2}} + 2gc{t_{\text{i}}} + 2f\frac{c}{{{t_{\text{i}}}}} + k = 0 \cr
& \Rightarrow {c^2}t_{\text{i}}^4 + 2gct_{\text{i}}^3 + kt_{\text{i}}^3 + 2fc{t_{\text{i}}} + {c^2} = 0 \cr} $$
Its roots are $${t_1},\,{t_2},\,{t_3},\,{t_4}$$ so $${t_1}{t_2}{t_3}{t_4} = \frac{{{c^2}}}{{{c^2}}} = 1$$
Also, $${t_1} + {t_2} + {t_3} + {t_4} = - \frac{{2gc}}{{{c^2}}} = - \frac{{2g}}{c}$$