Question

If $P(x_1,y_1),Q(x_2,y_2),R(x_3,y_3)$ and $S(x_4,y_4)$ are four concyclic points on the rectangular hyperbola $xy=c^2$, then coordinates of the orthocenter of the $△PQR$ is

• A. $(x_4,-y_4)$
• B. $(x_4,y_4)$
• C. $(-x_4,-y_4)$
• D. $(-x_4,y_4)$

Answer 1

The correct option is C $(-x_4,-y_4)$

Orthocentre of triangle formed by taking three points on rectangular hyperbola lies on same hyperbola.
let $(x_i,y_i)=(c{t}_i,\frac{c}{t_i})$ where $i=1,2,3,4$ represents given points

If $(c{t}_i,\frac{c}{t_i})$, where $i=1,2,3$ are the vertices of the triangle then the orthocentre is $(\frac{-c}{t_1{t}_2{t}_3},-c{t}_1{t}_2{t}_3)$,

Now let circle equation be
$x^2+y^2+2gx+2fy+r=0\Rightarrow x^2+\frac{c^4}{x^2}+2gx+\frac{2f{c}^2}{x}+r=0\Rightarrow x^4+2g{x}^3+r{x}^2+2f{c}^{2}x+c^4=0$
Absissa of intersection points are roots of above equation.
So, using product of roots
$x_1{x}_2{x}_3{x}_4=c^4$
$\Rightarrow t_1{t}_2{t}_3{t}_4=1$
Hence, orthocentre is $(-c{t}_4,\frac{-c}{t_4})=(-x_4,-y_4).$