Question

Find the equation of the hyperbola, if its asymptotes are parallel to $x+2y-12=0$ and $x-2y+8=0$. $(2,4)$ is the centre of the hyperbola and it passes through $(2,0)$.

Answer 1

Given the assymptotes are parallel to $x+2y-12=0$ and $x-2y+8=0$.
$\therefore$ the equation of the assymptotes are
$x+2y+l=0$ and $x-2y+m=0$
Also, the asymptotes pass through centre $(2,4)$ of the hyperbola
Hence $x+2y+l=0$
$\Rightarrow 2+2\left(4\right)+l=0$
$\Rightarrow 2+8+l=0$

$\Rightarrow l=-10$
$\Rightarrow x-2y+m=0$
$\Rightarrow 2-2\left(4\right)+m=0$
$\Rightarrow 2-8+m=0\Rightarrow m=6$
$\therefore$ Equations of the assymptotes are
$x+2y-10=0$ and $x-2y+6=0$
$\therefore$ Combined equation of the assymptotes are
$(x+2y-10)$ and $(x-2y+6)=0$
Equation of the hyperbola is of the form
$(x+2y-10)(x-2y+6)+k=0$
This passes through $(2,0)$
$(2+2(0)-10)(2-2(0)+6)+k=0$
$(-8)\left(8\right)+k=0\Rightarrow k=64$
Equation of the required hyperbola is
$(x+2y-10)(x-2y+6)+64=0$