The mid point of the chord 4 x-3 y=5 of the hyperbola 2 x^2-3...

Given, $4 x-3 y=5$ and $2 x^2-3 y^2=12\therefore \quad 2\left(\frac{5+3 y}{4}\right)^2-3 y^2=12$$\begin{aligned} & \Rightarrow \quad \frac{\left(25+9 y^2+30 y\right)}{8}-3 y^2=12 \\ & \Rightarrow \quad 15 y^2-30 y+71=0 \\ & \Rightarrow \quad y=\frac{30 \pm \sqrt{900-4260}}{30} \\ & =1 \pm \frac{\sqrt{-3360}}{30} \\ & \end{aligned}$$$Also,$$\begin{aligned} & 2 x^2-3\left(\frac{4 x-5}{3}\right)^2=12 \\ & \Rightarrow \quad 10 x^2-40 x+61=0 \\ & \end{aligned}$$\Rightarrow \quad x=\frac{40 \pm \sqrt{1600-4 \times 10 \times 61}}{2 \times 10}=\frac{40 \pm \sqrt{-840}}{20}=2 \pm \frac{\sqrt{-840}}{20}\therefore \quad$ Points are $A$\left(2+\frac{\sqrt{-840}}{20},1+\frac{\sqrt{-3360}}{30}\right)$and B\left(2-\frac{\sqrt{-840}}{20},1-\frac{\sqrt{-3360}}{30}\right) \text {. }\therefore \quad Mid point of$A B$is$(2,1)$.$