To derive the equation involving $x$ and $y$ that eliminates $t$, we are going to manipulate the given parametric equations step by step.

The parametric equations are:

$x(t) = at + \frac{b}{t}$ $y(t) = ct - \frac{d}{t}$

Step 1: Solve for $t$ from $x(t)$

We start with the equation for $x(t)$:

$x(t) = at + \frac{b}{t}$

Multiply both sides of this equation by $t$ to eliminate the denominator:

$xt = at^2 + b$

Rearrange this equation to get a quadratic form:

$at^2 - xt + b = 0$

Now, we can solve this quadratic equation for $t$ using the quadratic formula:

$t = \frac{-(-x) \pm \sqrt{(-x)^2 - 4(a)(b)}}{2a} = \frac{x \pm \sqrt{x^2 - 4ab}}{2a}$

This gives us two possible solutions for $t$:

$t = \frac{x + \sqrt{x^2 - 4ab}}{2a} \quad \text{or} \quad t = \frac{x - \sqrt{x^2 - 4ab}}{2a}$

For simplicity, we will use the positive root (since $t > 0$):

$t = \frac{x + \sqrt{x^2 - 4ab}}{2a}$

Step 2: Substitute $t$ into $y(t)$

Next, we substitute the expression for $t$ into the equation for $y(t)$:

$y(t) = ct - \frac{d}{t}$

Substitute the expression for $t$:

$y = c \left( \frac{x + \sqrt{x^2 - 4ab}}{2a} \right) - \frac{d}{\frac{x + \sqrt{x^2 - 4ab}}{2a}}$

Simplifying the second term:

$y = \frac{c(x + \sqrt{x^2 - 4ab})}{2a} - \frac{2ad}{x + \sqrt{x^2 - 4ab}}$

Step 3: Eliminate $t$ to form a relation between $x$ and $y$

At this point, the expression for $y$ involves $t$ in a very complicated form, and further simplifications using algebraic manipulations can be done to express the equation purely in terms of $x$ and $y$, but it typically involves more advanced steps such as squaring and clearing the square root terms, as well as expanding. The final result will be a polynomial equation that relates $x$ and $y$ and eliminates $t$.

Conclusion

Thus, the geometric relation between $x$ and $y$ will be a polynomial equation of some degree involving $a, b, c,$ and $d$, but the precise form will depend on detailed algebraic manipulations.

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Follow-up questions:

1. How can we simplify this equation further to eliminate the square root?
2. What happens if we use the other root for $t$?
3. How do we interpret the resulting polynomial equation geometrically?
4. Can we find specific values for $a, b, c,$ and $d$ to simplify the equation further?
5. How do parametric equations relate to polar coordinates?

Tip:

When working with parametric equations, always try to eliminate the parameter by expressing it in terms of the other variables. Use algebraic manipulation like substitution, solving for $t$, or even squaring to help eliminate $t$.