If t is a parameter then what does the following represent-
${\text{x}} = {\text{a}}\left( {{\text{t}} + \dfrac{1}{{\text{t}}}} \right),\;{\text{y}} = {\text{b}}\left( {{\text{t}} - \dfrac{1}{{\text{t}}}} \right)$
A. Ellipse
B. Circle
C. Pair of straight lines
D. Hyperbola

Hint: In such types of questions, we have to eliminate the parameter using general algebra and check the conditions for the final equation, which will tell us about the type of conic.

Complete step-by-step solution -
Now on multiplying the ${\text{x}} = {\text{a}}\left( {{\text{t}} + \dfrac{1}{{\text{t}}}} \right)$ by ‘b’ and ${\text{y}} = {\text{b}}\left( {{\text{t}} - \dfrac{1}{{\text{t}}}} \right)$ by ‘a’ we get,
$bx = ab\left( {{\text{t}} + \dfrac{1}{{\text{t}}}} \right)\;and\;ay = ab\left( {{\text{t}} - \dfrac{1}{{\text{t}}}} \right)$
Using the identity-
${\left( {{\text{a}} + {\text{b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2} + 2ab$
Squaring both sides,
$\begin{align}
  &{{\text{b}}^2}{{\text{x}}^2} = {{\text{a}}^2}{{\text{b}}^2}\left( {{{\text{t}}^2} + \dfrac{1}{{{{\text{t}}^2}}} + 1} \right)\;.............................\left( 1 \right) \\
  &{{\text{a}}^2}{{\text{y}}^2} = {{\text{a}}^2}{{\text{b}}^2}\left( {{{\text{t}}^2} + \dfrac{1}{{{{\text{t}}^2}}} - 1} \right)..................................\left( 2 \right) \\
\end{align} $
Subtracting equation (1) and (2),
$b^2x^2 - a^2y^2 = a^2b^2(2 - (-2))$
$b^2x^2 - a^2y^2 = 4a^2b^2$
Dividing both the sides by $4{a}^2{b}^2$,
$\dfrac{{{{\text{x}}^2}}}{{{{\text{(2a)}}^2}}} - \dfrac{{{{\text{y}}^2}}}{{{{\text{(2b)}}^2}}} = 1$

This is the equation of a hyperbola, hence the correct option is D. Hyperbola

Note: In such types of questions, we have to analyze the given parametric equations and try to eliminate the parameter using algebraic identities. For example, we multiplied x by b and y by a so that we get similar terms in the equations. Then we squared and subtracted to eliminate the parameter easily. The parameter can be eliminated by other methods as well, but it is advisable to analyze the equations to find the best possible method.