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 $(1,1)$,then combined equation of the asymptotes is

• A. $xy-1=x-y$
• B. $xy+1=x+y$
• C. $2xy-x+y=0$
• D. $2xy+x-y=0$

Answer 1

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

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
$(x-1)(y-1)=c^2$

As asymptotes are lines passing through centre and parallel to coordinate axes, combined equation of asymptotes is $(x-1)(y-1)=0$