Question

The asymptotes of a hyperbola having centre at the point $(3,-2)$ are parallel to the lines $2x-y=0$ and $x+2y=0$. If the hyperbola passes through the point $(1,3)$, Find the equation of the hyperbola.

Answer 1

Let the asymptotes be $2x-y+c_1=0$ and $x+2y+c_2=0$

Since, the asymptotes passes through the centre $(3,-2)$ of the hyperbola.

$\therefore ,6+2+c_1=0$ and $3-4+c_2=0$

$\Rightarrow c_1=-8,c_2=1$

Thus, the equations of the asymptotes are

$2x-y-8=0$ and $x+2y+1=0$

Let the equation of the hyperbola be

$(2x-y-8)(x+2y+1)+\lambda =0$ ........$\left(1\right)$

It passes through $(1,3)$.

$\therefore ,(2-3-8)(1+6+1)+\lambda =0$

$\Rightarrow -9\times 8+\lambda =0$

$\Rightarrow -72+\lambda =0$

$\Rightarrow \lambda =72$

Putting the value of $\lambda$ in $\left(1\right)$, we obtain

$(2x-y-8)(x+2y+1)+72=0$

This is the equation of the required hyperbola.