Answer:
Option B
Explanation:
We know that equation of normal of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is
$\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}$
$\therefore$ Equations of normal to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$
is
$\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=9+4$
$\Rightarrow$ $\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=13$
Since, line x+y=k is normal to the given hyperbola
$\therefore$ $\frac{\frac{9}{x_{1}}}{1}=\frac{\frac{4}{y_{1}}}{1}=\frac{13}{k}$
$\Rightarrow$ $\frac{9}{x_{1}}=\frac{4}{y_{1}}$
= $\frac{13}{k}$
$\Rightarrow$ $x_{1}= \frac{9k}{13}$
$y_{1}=\frac{4k}{13}$
Since ($(x_{1},y_{1})$ lie on the hyperbola
$\therefore$ $\frac{\left(\frac{9k}{13}\right)^{2}}{9}-\frac{\left(\frac{4k}{13}\right)^{2}}{9}=1$
$\Rightarrow$ $\frac{9k^{2}}{169}-\frac{4k^{2}}{169}=1$
$\Rightarrow$ $5k^{2}=169$
$\Rightarrow$ $k=\pm\frac{13}{\sqrt{5}}$
