Question

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

Answer 1

Let the asymptotes be $x-3y+c_1=0$ and $2x-y+c_2=0$

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

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

$\Rightarrow c_1=9,c_2=8$

Thus, the equations of the asymptotes are

$x-3y+9=0$ and $2x-y+8=0$

Let the equation of the hyperbola be

$(x-3y+9)(2x-y+8)+\lambda =0$ ........$\left(1\right)$

It passes through $(1,-3)$.

$\therefore ,(1+9+9)(2+3+8)+\lambda =0$

$\Rightarrow 19\times 13+\lambda =0$

$\Rightarrow 247+\lambda =0$

$\Rightarrow \lambda =-247$

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

$(x-3y+9)(2x-y+8)-247=0$

This is the equation of the required hyperbola.