Let's solve this step by step:

We are given the parametric equations: $x(t) = at + \frac{b}{t}, \quad y(t) = ct - \frac{d}{t},$ where $a, b, c, d$ are constants, and $t > 0$.

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Step 1: Solve for $t$ in terms of $x$ and $y$

Start by rearranging the equations to express $t$ in terms of $x$ and $y$.

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

Multiply both sides by $t$ to eliminate the fraction: $xt = at^2 + b.$ Rearranging gives: $at^2 - xt + b = 0.$ This is a quadratic equation in $t$.

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

Multiply both sides by $t$ similarly: $yt = ct^2 - d.$ Rearranging gives: $ct^2 - yt - d = 0.$ This is another quadratic equation in $t$.

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Step 2: Eliminate $t$ using the resultants of the quadratic equations

The general method to eliminate $t$ involves finding a resultant of two polynomials. Since both equations are quadratic in $t$, we can eliminate $t$ by finding the determinant of the coefficients of $t^2$, $t$, and the constant terms from both equations.

• For the equation $at^2 - xt + b = 0$, the coefficients are: $\text{Coefficients: } (a, -x, b).$
• For the equation $ct^2 - yt - d = 0$, the coefficients are: $\text{Coefficients: } (c, -y, -d).$

The resultant $R$ of two quadratic equations $A_1t^2 + B_1t + C_1 = 0$ and $A_2t^2 + B_2t + C_2 = 0$ is given by the determinant:

a & -x & b \\ c & -y & -d \\ 0 & 1 & 0 \end{vmatrix}.$$ --- ### **Step 3: Compute the determinant** Expanding the determinant: $$R = a \begin{vmatrix} -y & -d \\ 1 & 0 \end{vmatrix} - (-x) \begin{vmatrix} c & -d \\ 0 & 0 \end{vmatrix} + b \begin{vmatrix} c & -y \\ 0 & 1 \end{vmatrix}.$$ Expanding each of these $$2 \times 2$$ determinants: 1. $$\begin{vmatrix} -y & -d \\ 1 & 0 \end{vmatrix} = (0)(-y) - (-d)(1) = d,$$ so the first term becomes $$a \cdot d = ad$$. 2. $$\begin{vmatrix} c & -d \\ 0 & 0 \end{vmatrix} = 0,$$ so the second term is zero. 3. $$\begin{vmatrix} c & -y \\ 0 & 1 \end{vmatrix} = (c)(1) - (-y)(0) = c,$$ so the third term becomes $$b \cdot c = bc$$. Thus, the resultant is: $$R = ad + bc.$$ --- ### **Step 4: Final polynomial equation** The polynomial equation involving $$x$$ and $$y$$ that eliminates $$t$$ is: $$ad + bc = 0.$$ --- ### Would you like further explanation or simplification of the steps?