Question

If t is a parameter, then $x = a\left( {t + \dfrac{1}{t}} \right),y = b\left( {t - \dfrac{1}{t}} \right)$ represents which of the following?
(A). An ellipse
(B). A circle
(C). A pair of straight lines
(D). A hyperbola

Answer 1

Hint- In such types of questions, we will approach the problem in such a way that we will just take the two coordinates into consideration and transpose the a and b part to the left-hand side of the equations respectively. After that we will just take the square of the L.H.S and the R.H.S and just expand the R.H.S using the identities like ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$. Now, we will just arrange the two equations using the addition, subtraction rules to get to know they satisfy and match the equation of which conic.

Complete step-by-step answer:

We have been given that $x = a\left( {t + \dfrac{1}{t}} \right),y = b\left( {t - \dfrac{1}{t}} \right)$
Now, transpose the a and b to L.H.S in both the equations we get
$\dfrac{x}{a} = \left( {t + \dfrac{1}{t}} \right),\dfrac{y}{b} = \left( {t - \dfrac{1}{t}} \right)$
Take the square of L.H.S and R.H.S in both the equations.
${\left( {\dfrac{x}{a}} \right)^2} = {\left( {t + \dfrac{1}{t}} \right)^2},{\left( {\dfrac{y}{b}} \right)^2} = {\left( {t - \dfrac{1}{t}} \right)^2}$
We will just consider the first equation i.e. ${\left( {\dfrac{x}{a}} \right)^2} = {\left( {t + \dfrac{1}{t}} \right)^2}$
Using the identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$ on R.H.S we get,
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} = {(t)^2} + {\left( {\dfrac{1}{t}} \right)^2} + 2 \times t \times \dfrac{1}{t}$
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} = {(t)^2} + {\left( {\dfrac{1}{t}} \right)^2} + 2$ ----(1)
We will just consider the second equation i.e. ${\left( {\dfrac{y}{b}} \right)^2} = {\left( {t - \dfrac{1}{t}} \right)^2}$
Using the identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$ on R.H.S we get,
$\Rightarrow {\left( {\dfrac{y}{b}} \right)^2} = {(t)^2} + {\left( {\dfrac{1}{t}} \right)^2} - 2 \times t \times \dfrac{1}{t}$
$\Rightarrow {\left( {\dfrac{y}{b}} \right)^2} = {(t)^2} + {\left( {\dfrac{1}{t}} \right)^2} - 2$ ----(2)
Now, on subtracting (2) from (1) we get,
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} - {\left( {\dfrac{y}{b}} \right)^2} = {(t)^2} + {\left( {\dfrac{1}{t}} \right)^2} + 2 - \left( {{{(t)}^2} + {{\left( {\dfrac{1}{t}} \right)}^2} - 2} \right)$
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} - {\left( {\dfrac{y}{b}} \right)^2} = {(t)^2} + {\left( {\dfrac{1}{t}} \right)^2} + 2 - {(t)^2} - {\left( {\dfrac{1}{t}} \right)^2} + 2$
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} - {\left( {\dfrac{y}{b}} \right)^2} = 4$
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} - {\left( {\dfrac{y}{b}} \right)^2} = {(2)^2}$
Transposing ${(2)^2}$ to L.H.S we get,
$\Rightarrow \dfrac{1}{{{{(2)}^2}}}{\left( {\dfrac{x}{a}} \right)^2} - \dfrac{1}{{{{(2)}^2}}}{\left( {\dfrac{y}{b}} \right)^2} = 1$
$\Rightarrow \dfrac{{{x^2}}}{{4{a^2}}} - \dfrac{{{y^2}}}{{4{b^2}}} = 1$
Now, we know that the standard form of the equation of a hyperbola with center $(0,0)$ and transverse axis on the x-axis is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
Hence, we can say that the equation so formed in our question is of hyperbola. So, $x = a\left( {t + \dfrac{1}{t}} \right),y = b\left( {t - \dfrac{1}{t}} \right)$ represents a hyperbola.

$\therefore$ Option. D A hyperbola is our correct answer.

Note- For such types of question we must be knowing the standard form of the equation of a hyperbola with center $(0,0)$ and transverse axis on the x-axis which is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ so as to compare the equations and also we should be knowing the basic algebraic identities like ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$ . Moreover, we should be knowing the equations of all the conics to just check whether our answer is similar to which of the conics.