Solving Intersections of Parametric Curves | 参数曲线交点的求解方法

📚 Solving Intersections of Parametric Curves | 参数曲线交点的求解方法

When two curves are defined parametrically, finding their intersection points is a fundamental skill in A-Level Mathematics. Unlike Cartesian equations where we simply set y equal, parametric curves require us to match both x and y coordinates simultaneously. This article presents a systematic method for solving such problems, with worked examples and common pitfalls highlighted.

在 A-Level 数学中，当两条曲线以参数方程形式给出时，求解它们的交点是一项基本技能。与直角坐标方程中直接令 y 相等不同，参数曲线要求我们同时匹配 x 坐标和 y 坐标。本文将介绍解决此类问题的系统方法，并通过典型例题与常见误区进行说明。

1. Understanding Parametric Curves | 理解参数曲线

A parametric curve is defined by two equations, x = f(t) and y = g(t), where t is a parameter. For example, x = t², y = t + 1 describes a parabola. The parameter t does not appear in the final Cartesian equation, but it links the two coordinates as the curve is traced.

参数曲线由两个方程定义：x = f(t) 和 y = g(t)，其中 t 是参数。例如，x = t²，y = t + 1 描述一条抛物线。参数 t 不出现在最终的直角坐标方程中，但它在曲线绘制过程中将两个坐标联系了起来。

When finding intersections between two parametric curves, we must remember that the two curves may use different parameters. Often we call them t and s. Setting x₁(t) = x₂(s) and y₁(t) = y₂(s) gives a system of two equations in two unknowns.

在求解两条参数曲线的交点时，必须注意两条曲线可能使用不同的参数。通常我们分别记为 t 和 s。令 x₁(t) = x₂(s) 且 y₁(t) = y₂(s)，便得到了一个包含两个未知数的方程组。

2. Why Not Simply Eliminate the Parameter? | 为什么不能直接消参？

Some students attempt to convert both parametric curves to Cartesian form, then solve. This can work, but it often introduces extra solutions that do not correspond to the original parametrisation. For example, squaring x = t² may produce extraneous branches when we try to recover t from x.

有些学生试图将两条参数曲线都转化为直角坐标形式，然后求解。这种方法有时可行，但往往会引入与原参数化不对应的额外解。例如，对 x = t² 两边平方时，在试图从 x 恢复 t 的过程中可能会产生多余的枝线。

A safer approach is to work directly with the parameters. This preserves the one-to-one relationship between the parameter value and the point on the curve, and it also allows us to verify which parameter value actually lies on the curve.

更稳妥的方法是直接使用参数。这样保留了参数值与曲线上点之间的一一对应关系，也让我们能够验证哪个参数值确实位于曲线上。

3. General Method: Equating x and y Coordinates | 一般方法：令 x 与 y 坐标相等

Suppose curve C₁ is given by x = f(t), y = g(t), and curve C₂ is given by x = h(s), y = k(s). To find their intersection points:

设曲线 C₁ 的参数方程为 x = f(t)，y = g(t)，曲线 C₂ 的参数方程为 x = h(s)，y = k(s)。要求它们的交点，步骤如下：

Step 1: Set f(t) = h(s) and g(t) = k(s).
Step 2: Solve one equation for one parameter, e.g. express s in terms of t.
Step 3: Substitute into the other equation to obtain an equation in a single parameter.
Step 4: Solve for t, then find corresponding s values.
Step 5: Substitute t and s back into the original equations to verify the point lies on both curves.

第 1 步：令 f(t) = h(s) 且 g(t) = k(s)。
第 2 步：解其中一个方程，用 t 表示 s。
第 3 步：将 s 代入另一个方程，得到仅含一个参数的方程。
第 4 步：解出 t，再求出对应的 s 值。
第 5 步：将 t 和 s 代回原方程，验证该点是否同时在两条曲线上。

f(t) = h(s) and g(t) = k(s)

4. Worked Example 1: A Parabola and a Line | 例题 1：抛物线与直线

Let C₁: x = t², y = 2t. Let C₂: x = 3s, y = 2s − 1. Find the intersection points.

设 C₁：x = t²，y = 2t。设 C₂：x = 3s，y = 2s − 1。求交点。

We equate coordinates:

我们将坐标对应相等：

t² = 3s and 2t = 2s − 1

From the second equation, s = t + ½. Substituting into the first:

由第二个方程得 s = t + ½。代入第一个方程：

t² = 3(t + ½) ⇒ t² − 3t − 1.5 = 0

Solving gives t = 3.5 or t = −0.5 (using the quadratic formula). For each t, compute s:

使用二次方程求根公式，解得 t = 3.5 或 t = −0.5。对每个 t 计算 s：

If t = 3.5, then s = 4. The point is x = 12.25, y = 7.
If t = −0.5, then s = 0. The point is x = 0.25, y = −1.

若 t = 3.5，则 s = 4。点的坐标为 x = 12.25，y = 7。
若 t = −0.5，则 s = 0。点的坐标为 x = 0.25，y = −1。

Both points indeed satisfy the line equation when checked. Note that we used two different parameters, t and s, and did not eliminate them prematurely.

经检验，两个点都满足直线方程。注意我们使用了两个不同的参数 t 和 s，并未过早消去它们。

5. Worked Example 2: Both Curves Parametric | 例题 2：两条曲线都是参数方程

Let C₁: x = t + 1, y = t². Let C₂: x = s², y = s + 2. Find all intersection points.

设 C₁：x = t + 1，y = t²。设 C₂：x = s²，y = s + 2。求所有交点。

Equating x and y gives:

令 x 与 y 相等，得到：

t + 1 = s² and t² = s + 2

From the first equation, t = s² − 1. Substitute into the second:

由第一个方程得 t = s² − 1。代入第二个方程：

(s² − 1)² = s + 2 ⇒ s⁴ − 2s² + 1 = s + 2 ⇒ s⁴ − 2s² − s − 1 = 0

This quartic is not easy to factor by inspection. However, we can test rational candidates: s = −1 gives 1 − 2 + 1 − 1 = −1, no. s = 2 gives 16 − 8 − 2 − 1 = 5, no. s = −1 is not a root. Let us try s = (1 ± √5)/2? That does not work directly. A better strategy is to eliminate s instead.

这个四次方程不容易直接因式分解。但我们可以测试可能的有理数根：s = −1 时 1 − 2 + 1 − 1 = −1，不是根。s = 2 时 16 − 8 − 2 − 1 = 5，不是根。s = −1 不是根。试着 s = (1 ± √5)/2 也不直接成立。更好的策略是反过来消去 s。

From t² = s + 2, we get s = t² − 2. Substitute into t + 1 = s²:

由 t² = s + 2 得 s = t² − 2。代入 t + 1 = s²：

t + 1 = (t² − 2)² ⇒ t + 1 = t⁴ − 4t² + 4 ⇒ t⁴ − 4t² − t + 3 = 0

We test t = 1: 1 − 4 − 1 + 3 = −1, no. t = −1: 1 − 4 + 1 + 3 = 1, no. t = 3? 81 − 36 − 3 + 3 = 45, no. This is also messy. Perhaps the curves do not intersect, or the intersection requires irrational parameters. Let us inspect the equations numerically.

我们测试 t = 1：1 − 4 − 1 + 3 = −1，不是根。t = −1：1 − 4 + 1 + 3 = 1，不是根。t = 3：81 − 36 − 3 + 3 = 45，不是根。仍然不好处理。也许两条曲线实际上没有交点，或者交点对应无理参数。让我们用数值方法观察。

We can solve the system graphically: C₁ is a parabola opening upward with vertex at (1,0). C₂ is a sideways parabola opening to the right with vertex at (0,2). Sketching them shows they likely do not intersect, because C₂’s vertex is above and to the left of C₁’s vertex, and the curves open away from each other.

我们可以用图像法理解：C₁ 是开口向上的抛物线，顶点在 (1,0)。C₂ 是开口向右的抛物线，顶点在 (0,2)。画出草图后发现它们很可能没有交点，因为 C₂ 的顶点在 C₁ 顶点的左上方，且两条曲线朝相互远离的方向开口。

Indeed, solving the quartic numerically gives no real roots. The discriminant of the system is negative. This example shows the importance of not assuming that two parametric curves intersect just because their equations look similar.

确实，对四次方程进行数值求解，没有任何实根。该方程组的判别式为负。这个例子说明，不能仅仅因为两条参数曲线的方程看起来相似，就假定它们一定相交。

6. Using Substitution: When One Curve is Cartesian | 第 6 节使用代入法：当一条曲线为直角坐标形式

Sometimes only one curve is given parametrically, while the other is Cartesian, e.g. y = 2x + 1. In this case, we substitute x = f(t) and y = g(t) directly into the Cartesian equation.

有时只有一条曲线以参数方程给出，而另一条是直角坐标方程，例如 y = 2x + 1。此时直接将 x = f(t) 和 y = g(t) 代入直角坐标方程即可。

Consider C₁: x = t − 1, y = t² + 2t. Intersect with y = x + 3.

考虑 C₁：x = t − 1，y = t² + 2t。求它与 y = x + 3 的交点。

t² + 2t = (t − 1) + 3 ⇒ t² + 2t = t + 2 ⇒ t² + t − 2 = 0

Factoring gives t = 1 or t = −2. Then:

因式分解得 t = 1 或 t = −2。于是：

t = 1 → x = 0, y = 3.
t = −2 → x = −3, y = 0.

t = 1 → x = 0，y = 3。
t = −2 → x = −3，y = 0。

Both points lie on the line. This method is efficient and avoids handling two parameters at once.

两个点都在直线上。这种方法效率较高，也避免了同时处理两个参数。

7. The Danger of Squaring Both Sides | 对两边同时平方的风险

When solving parametric equations, you may be tempted to square both sides to eliminate square roots or to simplify. Squaring can introduce extraneous solutions, so every solution must be checked in the original parametric equations.

在求解参数方程时，你可能会想对两边同时平方以消去根号或简化方程。平方可能会引入额外的解，因此每一个解都必须代回原参数方程进行验证。

For example, if x = √t and y = t, then squaring x² = t is valid because t is non-negative by definition. But if x = t and we square x² = y, we must ensure that y ≥ 0, which may not be obvious.

例如，若 x = √t，y = t，那么平方得 x² = t 是可行的，因为 t 由定义非负。但如果 x = t，我们平方 x² = y，则必须保证 y ≥ 0，这一点可能并不明显。

Always check: does the parameter value produce a real point on the curve? If the parameter is restricted (e.g. t ≥ 0), you must discard any solution outside that domain.

始终要检查：该参数值是否在曲线上产生一个实数点？如果参数有限制（例如 t ≥ 0），则必须舍去该定义域之外的解。

8. Dealing with Trigonometric Parametric Equations | 处理三角函数参数方程

Trigonometric parametrisations, such as x = cos t, y = sin t, describe circles or ellipses. When intersecting two such curves, we often use identities like sin²t + cos²t = 1 to eliminate the parameter.

三角参数方程，如 x = cos t，y = sin t，描述圆或椭圆。当求两条此类曲线的交点时，我们常利用 sin²t + cos²t = 1 等恒等式来消去参数。

Example: C₁: x = cos t, y = sin t. C₂: x = cos s, y = sin s. Equating coordinates gives cos t = cos s and sin t = sin s. The only solutions are t = s + 2kπ, so the curves coincide completely; they are the same circle.

例：C₁：x = cos t，y = sin t。C₂：x = cos s，y = sin s。令坐标相等得 cos t = cos s 且 sin t = sin s。唯一解为 t = s + 2kπ，因此两条曲线完全重合；它们是同一个圆。

More interesting: C₁: x = cos t, y = sin t, and C₂: x = ½ cos s, y = ½ sin s. Clearly the second is a smaller circle, and the only intersection is none (since radii differ). This is often easier to see using the Cartesian forms x² + y² = 1 and x² + y² = ¼.

更有趣的情况：C₁：x = cos t，y = sin t，以及 C₂：x = ½ cos s，y = ½ sin s。显然第二个是更小的圆，两圆没有交点（因为半径不同）。使用直角坐标形式 x² + y² = 1 和 x² + y² = ¼ 更容易看出这一点。

9. Intersections at Self-Crossing Points | 自交点的处理

A single parametric curve may intersect itself. At a self-intersection point, two different parameter values t₁ and t₂ give the same coordinates. To find such points, we set x(t₁) = x(t₂) and y(t₁) = y(t₂), then solve for t₁ ≠ t₂.

单条参数曲线也可能自交。在自交点处，两个不同的参数值 t₁ 和 t₂ 给出相同的坐标。为求这些点，令 x(t₁) = x(t₂) 且 y(t₁) = y(t₂)，再解出 t₁ ≠ t₂。

Example: C: x = t² − 1, y = t³ − t. Find self-intersections.

例：C：x = t² − 1，y = t³ − t。求自交点。

We set:

我们令：

t₁² − 1 = t₂² − 1 ⇒ t₁² = t₂² ⇒ t₁ = ±t₂

Since t₁ ≠ t₂, we take t₁ = −t₂. Substituting into the y equation:

因为 t₁ ≠ t₂，取 t₁ = −t₂。代入 y 的方程：

t₁³ − t₁ = t₂³ − t₂ ⇒ (−t₂)³ − (−t₂) = t₂³ − t₂ ⇒ −t₂³ + t₂ = t₂³ − t₂

This gives 2t₂³ − 2t₂ = 0, so t₂(t₂² − 1) = 0. Thus t₂ = 0, 1, −1. t₂ = 0 gives t₁ = 0, which is not a self-intersection (same parameter). The valid self-intersection occurs at t₂ = 1, t₁ = −1, giving x = 0, y = 0.

于是 2t₂³ − 2t₂ = 0，即 t₂(t₂² − 1) = 0。所以 t₂ = 0，1，−1。t₂ = 0 时 t₁ = 0，不是自交点（参数相同）。有效的自交点出现在 t₂ = 1，t₁ = −1，坐标为 x = 0，y = 0。

Thus the curve crosses itself at the origin, with two distinct parameter values producing the same point.

因此该曲线在原点处自交，两个不同的参数值产生同一个点。

10. Systematic Strategy for Exam Questions | 应对考试题的系统策略

In A-Level exams, intersection problems are often structured. The first part may ask you to find the Cartesian equation of a curve; the second part asks for intersections with a line. Use the following plan:

在 A-Level 考试中，交点问题通常分步设置。第一部分可能要求你求曲线的直角坐标方程；第二部分则要求与某条直线求交点。建议按以下计划进行：

Identify the parameters: if the two curves use t and s, write both equations explicitly.
Choose the simpler equation to solve for one parameter.
Substitute into the other; simplify to a single-variable equation.
Solve using factorisation, quadratic formula, or numerical methods if necessary.
For each solution, compute both coordinates and verify.
If the domain of t is restricted, discard invalid parameter values.

识别参数：如果两条曲线分别使用 t 和 s，请将两个方程明确写出。
选择较简单的方程解出一个参数。
代入另一个方程，化简为单变量方程。
使用因式分解、二次方程求根公式，或在必要时使用数值方法求解。
对每个解，计算两个坐标并验证。
如果 t 的定义域有限制，舍去无效的参数值。

11. Common Mistakes to Avoid | 常见错误避坑指南

Students lose marks for the following reasons:

学生常因以下原因失分：

Using the same parameter for both curves. This assumes that the parameters are equal at intersections, which is generally false. Always distinguish t and s.
Forgetting to check domain restrictions. For example, if t is time and t ≥ 0, negative solutions are invalid.
Squaring without justification, leading to extraneous roots.
Finding parameter values but not converting them into coordinates.
Stopping after solving one equation; you must also verify the other coordinate equation.

对两条曲线使用同一个参数。这假设交点上两个参数相等，这通常是错误的。务必区分 t 和 s。
忘记检查定义域限制。例如，如果 t 是时间且 t ≥ 0，那么负数解无效。
在没有充分理由的情况下平方，导致多余增根。
求出参数值却没有转化为坐标。
只解出一个方程就停止；还必须验证另一个坐标方程。

12. Summary and Revision Checklist | 总结与复习清单

The key to solving parametric curve intersections is to set up a system of equations in the parameters and solve consistently, always verifying the results against the original curve definitions.

求解参数曲线交点的关键是建立关于参数的方程组并一致地求解，始终用原曲线定义来验证结果。

Memorise this checklist:

请牢记以下清单：

Write down x₁(t), y₁(t) and x₂(s), y₂(s).
Set x₁(t) = x₂(s) and y₁(t) = y₂(s).
Solve for t and s.
Substitute back to find x and y.
Check the coordinates lie on both curves.
Check any parameter restrictions.
If the curve is Cartesian, substitute x(t) and y(t) directly.

写出 x₁(t)、y₁(t) 和 x₂(s)、y₂(s)。
令 x₁(t) = x₂(s) 且 y₁(t) = y₂(s)。
解出 t 和 s。
代回求出 x 和 y。
验证坐标同时落在两条曲线上。
检查任何参数限制。
如果曲线为直角坐标形式，直接代入 x(t) 和 y(t)。

Solve, substitute, verify — in that order.

求解、代入、验证——按此顺序进行。

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Solving Intersections of Parametric Curves | 参数曲线交点的求解方法