Answer:
Option B
Explanation:
The equation of any normal
to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $a x \cos \phi+by \cot \phi=a^{2}+b^{2}$
$\Rightarrow$ $a x \cos \phi+by \cot \phi -(a^{2}+b^{2})=0....(i)$
The straight line lx+my-n=0 will be normal to the hyperbola
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ , then eq.(i) and lx+my-n=0 represent the same line,
$\therefore$ $\frac{a \cos \phi}{l}=\frac{b \cot \phi}{m}=\frac{a^{2}+b^{2}}{n}$
$\Rightarrow$ $\sec \phi= \frac{na}{l(a^{2}+b^{2})}$
and $\tan \phi = \frac{nb}{m(a^{2}+b^{2})}$
$\frac{n^{2}a^{2}}{l^{2}(a^{2}+b^{2})^{2}}-\frac{n^{2}b^{2}}{m^{2}(a^{2}+b^{2})^{2}}=1$
$[ \because \sec^{2} \phi-\tan^{2} \phi=1]$
$\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{n^{2}}$
But given equation of normal is
$\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{k}$
$\therefore$ $k=n^{2}$
