Question

The asymptotes of a hyperbola are parallel to lines $2x+3y=0$ and $3x+2y=0.$ The hyperbola has its centre at $(1,2)$ and it passes through $(5,3).$ Find its equation.

• A. $(2x+3y-8)(3y+2y-7)=154$
• B. $(2x+3y-7)(3y+2y-8)=154$
• C. $(2x+3y-7)(3y+2y-8)=127$
• D. $(2x+3y-8)(3y+2y-7)=127$

Answer 1

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

let the equation of asymtotes be $2x+3y=a$ and $3x+2y=b$
both asymtotes intersect at centre $(1,2)$
Therefore, $a=2+3\left(2\right)=8$ and $b=3+2\left(2\right)=7$
now, the equation of hyperbola is of the form $(2x+3y-8)(3x+2y-7)=k$
It passes through $(5,3)$
Therefore, $\left(2\right(5)+3(3)-8)\left(3\right(5)+2(3)-7)=k$
Thus $k=154$