Question

Let t be a variable quantity, find the locus of the point (x, y) when
$x=a(t+\frac{1}{t})$ and $y=a(t-\frac{1}{t}).$

Answer 1

Given: $x=a(t+\frac{1}{t})$ and $y=a(t-\frac{1}{t})$

To get the required locus, we are supposed to eliminate 't' from these equations.

$\Rightarrow$ $\frac{x}{a}=(t+\frac{1}{t})$ ........ $\left(i\right)$

$\Rightarrow$ $\frac{y}{a}=(t-\frac{1}{t})$ ........ $\left(ii\right)$

Adding the eqn $\left(i\right)$ & $\left(ii\right)$, we get

$\Rightarrow$ $\frac{x}{a}+\frac{y}{a}=2t$ ....... $\left(I\right)$

Subtracting eqn $\left(i\right)$ & $\left(ii\right)$ gives,

$\Rightarrow$ $\frac{x}{a}-\frac{y}{a}=\frac{2}{t}$ ....... $\left(II\right)$

Multiply Eqns I & II,

$\Rightarrow$ $\frac{(x+y)(x-y)}{a^2}=4$

$\Rightarrow$ $x^2-y^2=4a^2$

This is the required locus of the point (x,y).