The pole of the line $lx+my+n=0$ with respect to the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, is
(a) $\left( \dfrac{{{a}^{2}}l}{n},\dfrac{{{b}^{2}}m}{n} \right)$
(b) $\left( -\dfrac{{{a}^{2}}l}{n},\dfrac{{{b}^{2}}m}{n} \right)$
(c) $\left( \dfrac{{{a}^{2}}l}{n},-\dfrac{{{b}^{2}}m}{n} \right)$
(d) $\left( -\dfrac{{{a}^{2}}l}{n},-\dfrac{{{b}^{2}}m}{n} \right)$

Hint: Here, we will the definition of the pole of a line with respect to a hyperbola to find the coordinates of the pole of the line $lx+my+n=0$ with respect to the given hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. The equation of the polar of the hyperbola is given as $\dfrac{hx}{{{a}^{2}}}-\dfrac{ky}{{{b}^{2}}}=1$ , where (h, k) represents the pole.

Complete step-by-step answer:
Let us consider any point A having coordinates (h, k) which lie inside or outside a hyperbola. If we draw tangents on the hyperbola from this point A such that the tangents touch the hyperbola at points Q and R, then the locus of points of intersection of tangents at Q and R is called polar.
The point A is called the pole.
The equation of the polar is given as:
$\dfrac{hx}{{{a}^{2}}}-\dfrac{ky}{{{b}^{2}}}=1$ , where (h, k) is the pole and a and b are the semi major axis and semi minor axis respectively of the hyperbola.
 Since, the given equation of the hyperbola is:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1..............(1)$
Also, the equation of the given line is:
$lx+my+n=0............\left( 2 \right)$.
And, we have a general equation of polar as:
$\dfrac{hx}{{{a}^{2}}}-\dfrac{ky}{{{b}^{2}}}=1............\left( 3 \right)$
So, the equation (2) and equation (3) represent the same line, that is they represent the equation of the polar.
On solving equation (2) and (3) for h and k, we get:
$\begin{align}
  & \dfrac{\left( \dfrac{h}{{{a}^{2}}} \right)}{l}=\dfrac{\left( \dfrac{k}{{{b}^{2}}} \right)}{m}=\dfrac{1}{-n} \\
 & \dfrac{h}{{{a}^{2}}l}=\dfrac{-k}{{{b}^{2}}m}=\dfrac{-1}{n} \\
\end{align}$
So, we have $h=\dfrac{-{{a}^{2}}l}{n}$ and $k=\dfrac{{{b}^{2}}m}{n}$
So, pole of the line $lx+my+n=0$ with respect to the given hyperbola is $\left( \dfrac{-{{a}^{2}}l}{n},\dfrac{{{b}^{2}}m}{n} \right)$.
Hence, option (b) is the correct answer.


Note: Students should remember the definition and equation of pole and polar. Here chances of mistake is that even after knowing the definition and equation students may not understand the question, So, the observation that the equation (2) and (3) represent the same line is important.