Q66 / 120 · VITEEE 2009 · Mathematics
The mid point of the chord $4 x-3 y=5$ of the hyperbola $2 x^{2}-3 y^{2}=12$ is
- A$\left(0,-\frac{5}{3}\right)$
- Correct answerB$(2,1)$
- C$\left(\frac{5}{4}, 0\right)$
- D$\left(\frac{11}{4}, 2\right)$
Solution
Solving $4 x-3 y=5$ and $2 x^{2}-3 y^{2}=12$
$$
\begin{array}{l}
\therefore 2\left(\frac{5+3 y}{4}\right)^{2}-3 y^{2}=12 \\
\Rightarrow \frac{\left(25+9 y^{2}+30 y\right)}{8}-3 y^{2}=12 \\
\Rightarrow 15 y^{2}-30 y+71=0 \\
\Rightarrow y=\frac{30 \pm \sqrt{900-4260}}{30}=1 \pm \frac{\sqrt{-3360}}{30}
\end{array}
$$
Also, $2 x^{2}-3\left(\frac{4 x-5}{3}\right)^{2}=12$
$$
\Rightarrow 10 x^{2}-40 x+61=0
$$
$$
\Rightarrow x=\frac{40 \pm \sqrt{1600-4 \times 10 \times 61}}{2 \times 10}
$$
$=\frac{40 \pm \sqrt{-840}}{20}=2 \pm \frac{\sqrt{-840}}{20}$
$\therefore$ Points are $A\left(2+\frac{\sqrt{-840}}{20}, 1+\frac{\sqrt{-3360}}{30}\right)$
and $B\left(2-\frac{\sqrt{-840}}{20}, 1-\frac{\sqrt{-3360}}{30}\right)$.
$\therefore$ Mid point of $A B$ is $(2,1)$.
Hence, the correct answer is option: 'B'.