📚 Finding Intersection Points of Parametric Curves | A-Level数学：参数方程曲线的交点求解
In A-Level Mathematics, curves are often defined by parametric equations, where both x and y are expressed in terms of a third variable, usually t. Finding the points where two such curves intersect requires solving a system of equations that combines both parametric definitions. This topic tests your ability to manipulate equations, handle multiple parameters, and interpret solutions geometrically.
在A-Level数学中，曲线常以参数方程定义，即x和y都表示为第三个变量（通常为t）的函数。求两条此类曲线的交点，需要联立两个参数定义来解方程组。这一考点考察你对方程变形、多参数处理以及解的几何意义的理解能力。
1. Understanding Parametric Curves | 理解参数曲线
A parametric curve in the plane is written as x = f(t), y = g(t). As t varies over a given interval, the point (x, y) traces out a curve. Unlike Cartesian equations y = f(x), a parametric form can represent loops, vertical segments, and more complex paths.
平面参数曲线写作x = f(t)，y = g(t)。当t在给定区间内变化时，点(x, y)描绘出一条曲线。与笛卡尔方程y = f(x)不同，参数形式可以表示环、竖直线段以及更复杂的路径。
For example, the circle x = cos t, y = sin t traces a unit circle as t runs from 0 to 2π. Another curve may be given by x = t², y = t³.
例如，圆x = cos t，y = sin t在t从0到2π时描出一个单位圆。另一条曲线可设为x = t²，y = t³。
2. The Core Idea of Finding Intersections | 求交点的核心思想
To find the intersection points of two parametric curves, we need values of the parameters (say t for the first curve and s for the second) such that the x-coordinates are equal and the y-coordinates are equal simultaneously. This gives a system of two equations in two unknowns, t and s.
要求两条参数曲线的交点，我们需要找到第一条曲线的参数t和第二条曲线的参数s，使它们在x坐标和y坐标上同时相等。这就得到一个关于t和s的二元方程组。
x₁(t) = x₂(s), y₁(t) = y₂(s)
Solving this system yields the parameter pairs, from which the actual coordinates can be computed.
解此方程组得到参数对，进而计算出实际坐标。
3. Using Different Parameters for Different Curves | 为不同曲线使用不同参数
A common mistake is to assume that both curves use the same parameter value at an intersection. In general, the parameter t for curve C₁ and the parameter s for curve C₂ are independent. You must distinguish them clearly.
一个常见错误是假设两条曲线在交点处使用相同的参数值。一般来说，曲线C₁的参数t和曲线C₂的参数s是相互独立的。你必须明确区分它们。
For instance, if C₁: x = t, y = t² and C₂: x = s², y = s, setting t = s would incorrectly describe a different relationship. Instead, solve x: t = s² and y: t² = s.
例如，若C₁: x = t，y = t²，C₂: x = s²，y = s，令t = s会错误地描述另一种关系。正确做法是联立x: t = s²，y: t² = s。
4. Substitution Method | 代入法
When one equation is easy to solve for one parameter, substitute it into the other equation. From t = s² and t² = s, substitute t = s² into t² = s to get (s²)² = s, i.e. s⁴ = s.
当其中一个方程容易解出某个参数时，可将其代入另一个方程。由t = s²代入t² = s，得到(s²)² = s，即s⁴ = s。
Then s(s³ – 1) = 0, so s = 0 or s = 1. The corresponding t values are 0 and 1. The intersection points are (0,0) and (1,1).
于是s(s³ – 1) = 0，得s = 0或s = 1。对应的t值为0和1。交点坐标为(0,0)和(1,1)。
5. Solving with Elimination of Parameters | 消参求解
Sometimes it is easier to convert one or both parametric equations into Cartesian form before finding intersections. For example, C₁: x = t, y = t² is simply y = x². C₂: x = s², y = s gives x = y².
有时先将一个或两个参数方程化为笛卡尔形式再求交点会更简便。例如C₁: x = t，y = t²就是y = x²；C₂: x = s²，y = s可得x = y²。
Solving y = x² and x = y² gives the same results: substitute x = y² into y = x² gives y = y⁴, so y = 0 or y = 1. Thus points (0,0) and (1,1).
联立y = x²与x = y²可得相同结果：将x = y²代入y = x²得y = y⁴，故y = 0或y = 1。于是点为(0,0)和(1,1)。
6. Comparing the Two Approaches | 两种方法的比较
The parameter-substitution method works directly with the given equations and avoids the need to eliminate parameters. The Cartesian-elimination method may simplify the algebra but can sometimes introduce extraneous solutions or lose information about the parameter range.
参数代入法直接处理给定方程，无需消参。消参法可能简化代数运算，但有时会引入额外解或丢失参数范围的信息。
| Method | Advantage | Caution |
| Parameter substitution | Uses original parameter relations directly | May require solving higher-degree equations |
| Cartesian elimination | Often reduces to familiar curve equations | Check parameter domain restrictions |
In exams, either method is acceptable if clearly justified.
在考试中，只要论证清晰，两种方法均可接受。
7. Handling Trigonometric Parametric Equations | 处理三角函数参数方程
Trigonometric parameters often lead to periodic solutions. Consider C₁: x = cos t, y = sin t and C₂: x = 1 + cos s, y = sin s. We set cos t = 1 + cos s and sin t = sin s.
三角函数参数常导致周期解。考虑C₁: x = cos t，y = sin t，C₂: x = 1 + cos s，y = sin s。我们令cos t = 1 + cos s，sin t = sin s。
From sin t = sin s, either t = s + 2kπ or t = π – s + 2kπ. If t = s + 2kπ, then cos t = cos s, giving cos t = 1 + cos t, impossible. If t = π – s + 2kπ, then cos t = –cos s, so –cos s = 1 + cos s, hence cos s = –½.
由sin t = sin s，有t = s + 2kπ或t = π – s + 2kπ。若t = s + 2kπ，则cos t = cos s，代入得cos t = 1 + cos t，无解。若t = π – s + 2kπ，则cos t = –cos s，故–cos s = 1 + cos s，得cos s = –½。
Thus s = 2π/3 or 4π/3. The corresponding points are (–½, ±√3/2). These are the two intersections of two identical circles offset by 1 unit.
因此s = 2π/3或4π/3。对应点为(–½, ±√3/2)。这是两个圆心相距1的等圆的交点。
8. Intersection with a Cartesian Curve | 参数曲线与笛卡尔曲线的交点
If one curve is given in Cartesian form y = f(x) and the other is parametric x = x(t), y = y(t), substitute the parametric expressions into the Cartesian equation. This gives an equation in t only.
若一条曲线以笛卡尔形式y = f(x)给出，另一条为参数形式x = x(t)，y = y(t)，则将参数表达式代入笛卡尔方程，得到仅含t的方程。
For example, find where the line y = 2x – 1 meets the parabola x = t, y = t² – 3t + 1. Substitute: t² – 3t + 1 = 2t – 1 ⇒ t² – 5t + 2 = 0.
例如，求直线y = 2x – 1与抛物线x = t，y = t² – 3t + 1的交点。代入得t² – 3t + 1 = 2t – 1，即t² – 5t + 2 = 0。
Using the quadratic formula, t = (5 ± √17)/2. Each t gives one intersection point:
由求根公式，t = (5 ± √17)/2。每个t给出一个交点：
x = (5 ± √17)/2, y = 4 ± √17
9. Checking for Repeated Roots and Tangency | 检查重根与相切
When the resulting equation in the parameter has a repeated root, the curves are tangent at that intersection. For example, if solving gives a perfect square, the two curves touch without crossing.
当所得参数方程有重根时，两曲线在该交点处相切。例如，若解出完全平方形式，说明两曲线相切而非交叉。
This extra interpretation is often tested. Always examine the discriminant or factorised form to decide whether the root is distinct or repeated.
这一额外解释常被考查。务必检查判别式或因式分解形式，以判断根是单根还是重根。
10. Domain Restrictions and Extraneous Solutions | 定义域限制与增根
Parametric curves are often defined only for a restricted range of t. When solving, you must ensure that the parameter values obtained lie within the given domains. Otherwise, the algebraic point may not actually lie on the curve.
参数曲线通常只在t的某个受限区间内有定义。求解时，必须确保所得参数值位于给定范围内。否则，代数上得到的点实际上不在曲线上。
For instance, if C: x = t², y = t³ with 0 ≤ t ≤ 1 has no point corresponding to t = –1, even though the Cartesian equation y² = x³ would include such a point.
例如，若C: x = t²，y = t³且0 ≤ t ≤ 1，那么t = –1对应的点并不存在，尽管其笛卡尔方程y² = x³会包含该点。
11. Worked Example: Two Parametric Curves | 例题：两条参数曲线的交点
Let C₁: x = t², y = t – 1 and C₂: x = s + 1, y = s². Find their intersection points.
设C₁: x = t²，y = t – 1；C₂: x = s + 1，y = s²。求它们的交点。
Set the x-coordinates equal: t² = s + 1, so s = t² – 1. Set the y-coordinates equal: t – 1 = s². Substitute s: t – 1 = (t² – 1)².
令x坐标相等：t² = s + 1，故s = t² – 1。令y坐标相等：t – 1 = s²。代入s：t – 1 = (t² – 1)²。
Expand: t – 1 = t⁴ – 2t² + 1 ⇒ t⁴ – 2t² – t + 2 = 0. Factor: (t – 1)(t³ + t² – t – 2) = 0. Further factor gives (t – 1)(t – 2)(t² + 3t + 1)? Let us check carefully.
展开：t – 1 = t⁴ – 2t² + 1，即t⁴ – 2t² – t + 2 = 0。因式分解得(t – 1)(t³ + t² – t – 2) = 0。继续分解为(t – 1)(t – 2)(t² + 3t + 1)？我们仔细验证。
Actually t⁴ – 2t² – t + 2 factors as (t – 1)(t³ + t² – t – 2), and the cubic factors as (t – 2)(t² + 3t + 1). Check: (t – 2)(t² + 3t + 1) = t³ + 3t² + t – 2t² – 6t – 2 = t³ + t² – 5t – 2, which is not correct. Let us use a more reliable factorisation.
实际上t⁴ – 2t² – t + 2可分解为(t – 1)(t³ + t² – t – 2)。该三次式分解为(t – 2)(t² + 3t + 1)？验证：(t – 2)(t² + 3t + 1) = t³ + 3t² + t – 2t² – 6t – 2 = t³ + t² – 5t – 2，不正确。我们采用更可靠的分解。
Since t = 1 is a root, divide the quartic by (t – 1): t⁴ – 2t² – t + 2 = (t – 1)(t³ + t² – t – 2). Test t = 2 in the cubic: 8 + 4 – 2 – 2 = 8 ≠ 0, so t = 2 is not a root. Test t = –1: –1 + 1 + 1 – 2 = –1 ≠ 0. Test t = –2: –8 + 4 + 2 – 2 = –4 ≠ 0. Therefore the cubic has no simple rational roots; solving would involve messy values. For an A-Level example, we instead choose a cleaner problem.
由于t = 1是根，用(t – 1)除四次式得t⁴ – 2t² – t + 2 = (t – 1)(t³ + t² – t – 2)。在三次式中测试t = 2：8 + 4 – 2 – 2 = 8 ≠ 0，故t = 2不是根。测试t = –1：–1 + 1 + 1 – 2 = –1 ≠ 0；t = –2：–8 + 4 + 2 – 2 = –4 ≠ 0。因此三次式没有简单有理根；求解涉及复杂数值。对于A-Level例题，我们改为更简洁的问题。
Let us instead take C₁: x = t², y = t – 1 and C₂: x = s + 1, y = 2s – 3. Then t² = s + 1 and t – 1 = 2s – 3. From the second, s = (t + 2)/2. Substitute into the first: t² = (t + 2)/2 + 1 = (t + 4)/2 ⇒ 2t² – t – 4 = 0.
我们改设C₁: x = t²，y = t – 1；C₂: x = s + 1，y = 2s – 3。则t² = s + 1，t – 1 = 2s – 3。由第二个方程得s = (t + 2)/2。代入第一个方程：t² = (t + 2)/2 + 1 = (t + 4)/2，即2t² – t – 4 = 0。
Then t = (1 ± √33)/4. Each yields an intersection point. This demonstrates a typical algebraic path.
于是t = (1 ± √33)/4。每个t对应一个交点。这展示了典型的代数求解过程。
12. Common Pitfalls and Exam Tips | 常见错误与考试技巧
First, never forget to use independent parameters. Second, always check the domain of the parameter for each curve. Third, after finding parameter values, substitute back to obtain coordinates, and verify that the coordinates satisfy both original equations.
第一，切勿忘记使用独立参数。第二，始终检查每条曲线的参数定义域。第三，求出参数值后回代得到坐标，并验证坐标满足两个原始方程。
Finally, if the problem asks for “the number of intersections”, consider whether repeated roots correspond to tangencies and whether periodic trigonometric parameters give multiple points in the required interval.
最后，若题目要求“交点的个数”，需考虑重根对应相切，以及周期性三角函数参数在指定区间内是否给出多个点。
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