Question

Find the equations of the rectangular hyperbola which has for one its asymtotes the line $x+2y-5=0$ and passes through the points $(6,0)$ and $(-3,0)$.

Answer 1

One aysmptote is $x+2y-5=0$

Any line perpendicular to this is $2x-y+k=0$

$\therefore$ combined equation of the asymptote is $(x+2y-5)(2x-y+k)=0$

Hence the equation of the hyperbola is $(x+2y-5)(2x-y+k)=k^{\prime }$

This passes through $(6,0)$ and $(-3,0)$

$\therefore \left(1\right)(12+k)=k^{\prime }and(-8)(-6+k)=k^{\prime }$

$\Rightarrow 12+k=k^{\prime }and48-8k=k^{\prime }$

Solving for $k$ and $k^{\prime }$ we get $k=4,k^{\prime }=16$

Thus the required equation of the rectangular hyperbola is $(x+2y-5)(2x-y+4)=16$