The equation of the hyperbola whose asymptotes are the lines $3 x+4 y-2=0$, $2 x+y+1=0$ and which passes…
The equation of the hyperbola whose asymptotes are the lines $3 x+4 y-2=0$, $2 x+y+1=0$ and which passes through the point $(1,1)$ is
$6 x^2+11 x y+4 y^2-30 x+2 y+7=0$
$6 x^2+11 x y+4 y^2-x+2 y-22=0$
$6 x^2+11 x y+4 y^2-x+2 y+22=0$
$6 x^2+11 x y+4 y^2-3 x-7 y-11=0$
Solution
Equation to the asymptotes are given as
$
\begin{aligned}
3 x+4 y-2 & =0 \\
2 x+y+1 & =0
\end{aligned}
$
and
Eqs.(i) and (ii) may be given by
$
(3 x+4 y-2)(2 x+y+1)=0
$
As, the equation to the hyperbola will differ from Eq. (iii) only by a constant, it may be given by
$
(3 x+4 y-2(2 x+y+1)=\lambda
$
(where $\lambda$ is a constant)
$(1,1)$ lies on the curve given by
Eq. (iv), we have
$
\begin{aligned}
& & (3+4-2)(2+1+1) & =\lambda \\
\Rightarrow & & (5)(4)=\lambda \Rightarrow \lambda & =20
\end{aligned}
$
Hence, the equation of the hyperbola will be
$
\begin{aligned}
& (3 x+4 y-2)(2 x+y+1)=20 \\
\Rightarrow \quad & \quad 6 x^2+3 x y+3 x+8 x y+4 y^2 \\
& \quad+4 y-4 x-2 y-2=20 \\
\Rightarrow \quad 6 x^2+4 y^2+11 x y-x+2 y-22=0 & \\
\Rightarrow \quad & 6 x^2+11 x y+4 y^2-x+2 y-22=0 .
\end{aligned}
$