Question

The asymptotes of a hyperbola have center at the point $(1,2)$ and are parallel to the lines $2x+3y=0$ and $3x+2y=0$. If the hyperbola passes through the point $(5,3)$, then its equation is

• A. $(3x+2y-8)(2x+3y+7)-338=0$
• B. $(2x+3y-8)(3x+2y-7)-154=0$
• C. $(3x+2y-8)(2x+3y-7)-156=0$
• D. $(2x+3y-7)(3x+2y+8)-348=0$

Answer 1

The correct option is B $(2x+3y-8)(3x+2y-7)-154=0$

The equation of asymptotes parallel to the lines $2x+3y=0$ and $3x+2y=0$ are
$2x+3y+\lambda =0\cdots \left(1\right)$
$3x+2y+\mu =0\cdots \left(2\right)$
It is given that these asmptotes pass through the center $(1,2).$
Substituting the point $(1,2)$ in $\left(1\right)$ and $\left(2\right)$, we get
$\lambda =-8.\mu =-7$

Thus, the combined equation of asymptotes is $(2x+3y-8)(3x+2y-7)=0$

Let the equation of the hyperbola be $(2x+3y-8)(3x+2y-7)+\omega =0$
This hyperbola passes through $(5,3)$
$\Rightarrow (2\cdot 5+3\cdot 3-8)(3\cdot 5+2\cdot 3-7)+\omega =0$
$\Rightarrow \omega =-154$

Hence, the equation of the hyperbola is $(2x+3y-8)(3x+2y-7)-154=0$