Question

The asymptotes of a hyperbola having centre at the point $(1,2)$ are parallel to the lines $2x+3y=0$ and $3x+2y=0$. If the hyperbola passes through the point $(5,3)$, show that is equation is $(2x+3y–8)(3x+2y+7)=154$.

Answer 1

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

Since, the asymptotes passes through the centre $(1,2)$ of the hyperbola.

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

$\Rightarrow c_1=-8,c_2=-7$

Thus, the equations of the asymptotes are

$2x+3y–8=0$ and $3x+2y–7=0$

Let the equation of the hyperbola be

$(2x+3y-8)(3x+2y-7)+\lambda =0$ ........$\left(1\right)$

It passes through $(5,3)$.

$\therefore ,(10+9-8)(15+6-7)+\lambda =0$

$\Rightarrow ,11\times 14+\lambda =0$

$\Rightarrow \lambda =-154$

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

$(2x+3y–8)(3x+2y–7)–154=0$

This is the equation of the required hyperbola.