Question

Find the equation to the hyperbola, whose asymptotes are the straight lines $x+2y+3=0$, and $3x+4y+5=0$, and which passes through the point $(1,-1)$.
Write down also the equation to the conjugate hyperbola.

Answer 1

Equation of a hyperbola and its asymptotes differ in constant term only.

$\therefore$ equation of the hyperbola with asymptotes $x+2y+3=0$ and $3x+4y+5=0$ can be represented as $(x+2y+3)(3x+4y+5)+\lambda =0$

We know that point $(1,-1)$ lies on the hyperbola

$\therefore (1+2\times (-1)+3)(3\times 1+4\times (-1)+5)+\lambda =0⟹\lambda =-8$

Substituting value of $\lambda$ , equation of the hyperbola is

$(x+2y+3)(3x+4y+5)-8=0$ or $3x^2+8y^2+10xy+14x+22y+7=0$

We know that $H+C=2A$ or $C=2A-H$, where H is Equation of Hyperbola, C is its conjugate and A is the asymptotes

Solving the above, we can get the equation of the conjugate as $3x^2+8y^2+10xy+14x+22y+23=0$