Question

The vertices of $△ABC$ lie on a rectangular hyperbola such that the orthocentre of the triangle is $(0,-1)$ and the asymptotes of the rectangular hyperbola are parallel to the co-ordinate axes. The two perpendicular tangents of the hyperbola intersect at the point $(2,3)$. Then which of the following point(s) lie on the hyperbola?

• A. $(3,11)$
• B. $(10,4)$
• C. $(4,7)$
• D. $(6,6)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(3,11)$
B $(10,4)$
C $(4,7)$

Let the center of the hyperbola be $(h,k)$. So,the axes are $y=h$ and $x=k$

The equation of the hyperbola is $(x-h)(y-k)=p$
As points $A,B\&C$ lie on the hyperbola ,the orthocenter of the $△ABC$ also lies on the same hyperbola
$(0-h)(-1-k)=p......\left(1\right)$
In rectangular hyperbola , the locus of point from which the perpendicular tangents are drawn is center .
$\to (h,k)=(2,3)\Rightarrow h=2,k=3$
From eq. $\left(1\right)\Rightarrow p=(-2)(-4)=8$
The eq. of hyperbola is $(x-2)(y-3)=8$