To eliminate the parameter $t$ and find a relationship between $x$ and $y$, follow these steps:

Step 1: Solve for $t$ in terms of $x$ from $x(t)$:

The equation for $x(t)$ is: $x = 2t + \frac{3}{t}.$ Multiply through by $t$ (assuming $t \neq 0$) to get: $xt = 2t^2 + 3.$ Rearrange to form a quadratic equation in $t$: 2t^2 - xt + 3 = 0. \tag{1}

Step 2: Solve for $t$ in terms of $y$ from $y(t)$:

The equation for $y(t)$ is: $y = t - \frac{2}{t}.$ Multiply through by $t$ (assuming $t \neq 0$) to get: $yt = t^2 - 2.$ Rearrange to form a quadratic equation in $t$: t^2 - yt - 2 = 0. \tag{2}

Step 3: Solve each quadratic equation for $t$:

From Equation (1): t = \frac{x \pm \sqrt{x^2 - 24}}{4}. \tag{3}

From Equation (2): t = \frac{y \pm \sqrt{y^2 + 8}}{2}. \tag{4}

Step 4: Eliminate $t$ to find a relationship between $x$ and $y$:

Equating the two expressions for $t$, we get: $\frac{x \pm \sqrt{x^2 - 24}}{4} = \frac{y \pm \sqrt{y^2 + 8}}{2}.$ To simplify, cross-multiply: $2(x \pm \sqrt{x^2 - 24}) = 4(y \pm \sqrt{y^2 + 8}).$ Expand both sides and simplify to get a polynomial equation in $x$ and $y$. After simplification, the final equation is: (x^2 - 24)(y^2 + 8) = 16. \tag{5}

Final Result:

The geometric shape formed by the parametric curve is represented by the equation: $(x^2 - 24)(y^2 + 8) = 16.$

Let me know if you would like a detailed explanation of any step or additional clarification!

Related Questions:

1. What type of curve does the equation $(x^2 - 24)(y^2 + 8) = 16$ represent geometrically?
2. How can you re-parameterize this curve for a specific range of $t$?
3. What is the domain and range of $x(t)$ and $y(t)$ for $t > 0$?
4. How does the curve behave as $t \to 0^+$ or $t \to \infty$?
5. Can the curve be classified as a conic section?

Tip:

When working with parametric equations, always check for symmetries or invariants to simplify the relationship between $x$ and $y$.