The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4 x+3 y-7=0$…
The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4 x+3 y-7=0$ and $x-2 y-1=0$ is
$4 x^2+5 x y-6 y^2-11 x+11 y+50=0$
$4 x^2+5 x y-6 y^2-11 x+11 y-43=0$
$4 x^2-5 x y-6 y^2-11 x+11 y+57=0$
$x^2-5 x y-y^2-11 x+11 y-43=0$
Solution
Since, the equation of the hyperbola differs from that of the joint equations of the tangents by a constant, therefore the equation of the hyperbola will be of the form
$(4 x+3 y-7)(x-2 y-1)+k=0$ ...(i)
$\therefore$ Since, the hyperbola passes through the point $(2,3)$.
$\Rightarrow \quad(8+9-7)(2-6-1)+k=0$
$\Rightarrow \quad(10)(-5)+k=0,-50+k=0$
$\Rightarrow \quad k=50$
Hence, from Eq. (i)
$(4 x+3 y-7)(x-2 y-1)+50=0$
$4 x^2+3 x y-7 x-8 x y-6 y^2+14 y-4 x$ $-3 y+7+50=0$
$4 x^2-5 x y-6 y^2-11 x+11 y+57=0$