基于两个含平方根的方程，用指定变量表示$x_p$和$y_p$的求解问询

Hey there! Let's work through this problem together—you're dealing with two ellipse equations (each one defines points where the sum of distances to two fixed foci is constant, which is the core definition of an ellipse). Squaring directly can get messy fast, so let's take a structured approach to simplify and solve:

1. Convert each ellipse equation to a quadratic algebraic form #

Let's start with the first equation:
$$\sqrt{(x_p-x_{a_1})^{2+(y_p-y_{a_1})}2}+\sqrt{(x_p-x_{b_1})^{2+(y_p-y_{b_1})}2}=D_1$$

First, isolate one radical to one side:
$$\sqrt{(x_p-x_{a_1})^{2+(y_p-y_{a_1})}2} = D_1 - \sqrt{(x_p-x_{b_1})^{2+(y_p-y_{b_1})}2}$$

Square both sides to eliminate the left radical:
$$(x_p-x_{a_1})^{2+(y_p-y_{a_1})}2 = D_1^2 - 2D_1\sqrt{(x_p-x_{b_1})^{2+(y_p-y_{b_1})}2} + (x_p-x_{b_1})^{2+(y_p-y_{b_1})}2$$

Now, move all non-radical terms to the left side and simplify using the difference of squares:
$$(x_p-x_{a_1})^2 - (x_p-x_{b_1})^2 + (y_p-y_{a_1})^2 - (y_p-y_{b_1})^2 - D_1^2 = -2D_1\sqrt{(x_p-x_{b_1})^{2+(y_p-y_{b_1})}2}$$

Expanding the left side (the squared terms will cancel out, leaving linear terms in $x_p$ and $y_p$):
$$(x_{b_1}-x_{a_1})(2x_p - x_{a_1}-x_{b_1}) + (y_{b_1}-y_{a_1})(2y_p - y_{a_1}-y_{b_1}) - D_1^2 = -2D_1\sqrt{(x_p-x_{b_1})^{2+(y_p-y_{b_1})}2}$$

Let's call the left-hand side $L_1(x_p, y_p)$ for simplicity. Now square both sides again to eliminate the remaining radical:
$$L_1(x_p, y_p)^2 = 4D_1^{2\left[(x_p-x_{b_1})}2+(y_p-y_{b_1})^2\right]$$

Expand and rearrange this into standard quadratic form:
$$A_1x_p^2 + B_1x_py_p + C_1y_p^2 + E_1x_p + F_1y_p + G_1 = 0$$

Repeat this entire process for the second equation to get its quadratic form:
$$A_2x_p^2 + B_2x_py_p + C_2y_p^2 + E_2x_p + F_2y_p + G_2 = 0$$

2. Solve the system of quadratic equations #

Now you have two quadratic equations. Here's how to solve them:

• Subtract a multiple of one quadratic equation from the other to eliminate either the $x_p^2$ or $y_p^2$ term. This will give you a linear equation in $x_p$ and $y_p$ (since the quadratic terms cancel out).
• Solve the linear equation for one variable in terms of the other (e.g., $y_p = mx_p + b$).
• Substitute this expression into one of the original quadratic equations. You'll get a single-variable quadratic equation in $x_p$ (or $y_p$), which you can solve using the quadratic formula.
• Plug the solutions back into the linear equation to find the corresponding values of the other variable.

3. Verify your solutions #

Since we squared the equations twice, it's crucial to plug each solution back into the original radical equations to eliminate any extraneous roots that might have been introduced during the squaring steps.

内容的提问来源于stack exchange，提问作者Hassen Dhia