Question

The vertices of $△ABC$ lie on a rectangular hyperbola such that the orthocentre of the triangle is $(3,2)$ and the asymptotes of the rectangular hyperbola are parallel to the coordinate axes. If two perpendicular tangents of the hyperbola intersect at the point $(1,1)$ ,theb equation of the rectangular hyperbola is

• A. $xy=2x+y-2$
• B. $2xy=x+2y+5$
• C. $xy=x+y+1$
• D. None of these

Answer 1

The correct option is C $xy=x+y+1$

Let the centre of rectangular hyperbola be$(h,k)$
then equation of the hyperbola is
$(x-h)(y-k)=c^2$

Since perpendicular tangents intersect at the centre of rectangular hyperbola

Hence, centre of hyperbola is $(1,1)$ and equation of the hyperbola will be
$(x-1)(y-1)=c^2$
$\because$ for any triangle whose vertices lie on rectangular hyperbola, orthocentre also lie on the same hyperbola
$\therefore (3,2)$ lies on the hyperbola
Hence required equation will be
$(x-1)(y-1)=2$