The mid point of the chord $4 x-3 y=5$ of the hyperbola $2 x^2-3 y^2=12$ is

The mid point of the chord $4 x-3 y=5$ of the hyperbola $2 x^2-3 y^2=12$ is
  1. $\left(0,-\frac{5}{3}\right)$
  2. $(2,1)$
  3. $\left(\frac{5}{4}, 0\right)$
  4. $\left(\frac{11}{4}, 2\right)$

Solution

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 $(2,1)$.

Asked in: AP EAMCET 2009

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