Question

If t is a parameter , then $x = a\left( {t + \dfrac{1}{t}} \right),y = b\left( {t - \dfrac{1}{t}} \right)$ represents
(a). An ellipse
(b). A circle
(c). A pair of straight lines
(d). A hyperbola

Answer 1

Hint: In this particular type of question we need to rearrange and make equations from the given information. Then we need to square the equations and subtract them to eliminate t. Finally we have compared the resultant equation to get its relation with the above options.

Complete step-by-step answer:
 $x = a\left( {t + \dfrac{1}{t}} \right),y = b\left( {t - \dfrac{1}{t}} \right)$
We need to eliminate t and find the relation between x and y .
$x = a\left( {t + \dfrac{1}{t}} \right) \Rightarrow \dfrac{x}{a} = t + \dfrac{1}{t}$
$y = b\left( {t - \dfrac{1}{t}} \right) \Rightarrow \dfrac{y}{b} = t - \dfrac{1}{t}$
Squaring both the above equations
${\left( {\dfrac{x}{a}} \right)^2} = {\left( {t + \dfrac{1}{t}} \right)^2} \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} = {t^2} + \dfrac{1}{{{t^2}}} + 2$ (i) ( since ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$ )
\[{\left( {\dfrac{y}{b}} \right)^2} = {\left( {t - \dfrac{1}{t}} \right)^2} \Rightarrow \dfrac{{{y^2}}}{{{b^2}}} = {t^2} + \dfrac{1}{{{t^2}}} - 2\] (ii)
Subtracting (ii) from (i)
$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 4$
$ \Rightarrow \dfrac{{{x^2}}}{{{{\left( {2a} \right)}^2}}} - \dfrac{{{y^2}}}{{{{\left( {2b} \right)}^2}}} = 1$
Which is the general equation of hyperbola ( $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ )

Note: Remember to recall the basic formula of hyperbola and other conic sections to compare with the final equation we get . Note that the equation left should be free from any third variable . Understand the step by step procedure while solving such types of questions .