Q66 / 120 · VITEEE 2019 · Mathematics
The equation of the chord of the hyperbola $25 x^{2}-16 y^{2}=400$ that is bisected at point $(5,3)$ is:
- A$135 x-48 y=481$
- Correct answerB$125 x-48 y=481$
- C$125 x-4 y=48$
- DNone of these
Solution
The equation of the chord, having mid-point as $\left(x_{1}, y_{1}\right)$, of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by
$T=S_{1} ......(i)$
where, $T=\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}-1$ and
$$
S_{1}=\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}-1
$$
According to the question,
$\left(x_{1}, y_{1}\right)=(5,3)$ and $a^{2}=16, b^{2}=25$ as $25 x^{2}-16 y^{2}=400$
$$
\begin{array}{l}
\Rightarrow \frac{x^{2}}{16}-\frac{y^{2}}{25}=1 \therefore \frac{5 x}{16}-\frac{3 y}{25}=\frac{25}{16}-\frac{9}{25}
\end{array}
$$
[Using (i)]
$$
\begin{array}{l}
\Rightarrow \quad 125 x-48 y=625-144 \Rightarrow 125 x-48 y=481
\end{array}
$$
Hence, the correct answer is option: 'B'.