📚 Points of intersection | 交点
Finding points of intersection between curves is one of the most fundamental skills in A‑Level Mathematics. Whether you are dealing with lines, parabolas, circles, or parametric curves, the ability to locate where two graphs meet unlocks a deeper understanding of geometry, algebra, and real‑world modelling. This article will guide you through the systematic methods needed to solve intersection problems, from simple simultaneous equations to the subtleties of discriminant analysis, all tailored to the Edexcel specification.
求曲线之间的交点是 A‑Level 数学中最基本的技能之一。无论面对直线、抛物线、圆还是参数曲线，能够准确定位两条图像的交点，可以让你更深入地理解几何、代数以及现实世界建模。本文将系统地引导你掌握求解交点问题的方法，从简单的联立方程到判别式分析的微妙之处，所有内容均紧扣 Edexcel 考纲。
1. What are points of intersection? | 什么是交点？
two curves cross or touch. For any two curves C₁: y = f(x) and C₂: y = g(x), the intersection points are the real solutions of f(x) = g(x). The same principle applies if the curves are given in implicit form, such as x² + y² = 25 and y = 3x + 1, or in parametric form.
交点是指同时满足两条曲线方程的坐标对 (x, y)。从图像上看，就是两条曲线相交或相切的位置。对于任意两条曲线 C₁: y = f(x) 和 C₂: y = g(x)，交点就是方程 f(x) = g(x) 的实数解。如果曲线以隐式形式给出，例如 x² + y² = 25 与 y = 3x + 1，或参数形式给出，原理也是一样的。
You must then find the corresponding other coordinate. If both curves are expressed as y = … may require careful manipulation to avoid losing solutions or introducing extraneous ones.
求交点通常会归结为经过代换后求解一个单变量方程，然后再求出相应的另一个坐标。如果两条曲线都表示为 y = … 的形式，直接将它们设为相等就是最直接的方法。如果其中一条曲线以隐式形式给出，代换依然是关键，但需要仔细处理方程，避免漏解或产生增根。
2. Setting up simultaneous equations | 建立联立方程
have the line y = 2x + 3 and the parabola y = x² − 4x + 1. The simultaneous equations are:
要找到两条曲线的交点，你需要联立求解它们的方程。假设有直线 y = 2x + 3 和抛物线 y = x² − 4x + 1，联立方程为：
y = 2x + 3 and y = x² − 4x + 1
By equating the right‑ 4x + 1 = 2x + 3, which simplifies to x² − 6x − 2 = 0. Solving this quadratic yields the x‑coordinates of the intersection points. Once the x‑ ck into the simpler original equation (usually the linear one) to find the corresponding y‑coordinates.
将等号右侧设为相等，就得到只含 x 的方程：x² − 4x + 1 = 2x + 3，化简为 x² − 6x − 2 = 0。解这个二次方程即可得到交点的 x 坐标。求出 x 值后，代入较简单的那条原始方程（通常是直线方程）便可求得相应的 y 坐标。
ute into x² + y² = r², which can lead to a quadratic in x and then a quadratic in y, but the principle remains: eliminate one variable and solve.
这种代换方法适用于任何可以轻松分离出一个变量的方程组。对于圆或椭圆，你可能需要代入 x² + y² = r²，这样会得到关于 x 的二次方程，然后再求 y，但原理保持不变：消去一个变量再求解。
3. Using the substitution method step by step | 逐步使用代入法
Let’s formalise the substitution method for curves given as y = f(x) and y = g(x).
我们对表示为 y = f(x) 和 y = g(x) 的曲线，规范一下代入法步骤。
Step 1: Set f(x) = g(x).
2: Rearrange the equation into standard form (e.g., ax² + bx + c = 0 if it is quadratic).
Solve for x using factorisation, completing the square, or the quadratic formula x = [−b ± √b² − 4ac] / 2a.
Step 4: For each real x found, plug it back into the simpler expression for y to get the full coordinate.
Step 5: Write the intersection points as ordered pairs (x, y).
步骤 1：设 f(x) = g(x)。
步骤 2：将方程化为标准形式（例如，如果是二次的，则化为 ax² + bx + c = 0）。
步骤 3：通过因式分解、配方法或求根公式 x = [−b ± √b² − 4ac] / 2a 求解 x。
步骤 4：对每一个求得的实数 x，代回较简单的 y 表达式，得到完整坐标。
步骤 5：将交点写成有序对 (x, y) 的形式。
Always check the domain of the original functions. For example, if f(x) involves a square root, you must ensure the radicand is non‑negative and reject any solutions that fall outside the domain.
务必检查原函数的定义域。例如，如果 f(x) 包含平方根，就必须确保被开方数非负，并舍去任何落在定义域之外的解。
Example: y = √x − 1 and y = x − 3 → √x − 1 = x − 3
r squaring and solving, you might obtain x = 5 or x = 2. However, substituting x = 2 gives √1 = 2 − 3 which is −1, false. So only x = 5 is valid.
经过平方并求解后，你可能会得到 x = 5 或 x = 2。然而，将 x = 2 代入得 √1 = 2 − 3，即 1 = −1，不成立，因此只有 x = 5 是有效的解。
4. The elimination method for linear‑quadratic systems | 线性‑二次方程组的消元法
When one equation is linear and the other is quadratic, substitution is usually straightforward. However, sometimes it is convenient to use elimination—especially if both equations are given in general form. For instance, solve the system:
当一个方程是线性的、另一个是二次的，代入法通常很简单。但有时使用消元法也很方便——尤其是当两个方程都以一般形式给出时。例如，求解方程组：
2x + y = 7 and x² + y² = 25
simplify to obtain a quadratic in x. This is essentially substitution; true elimination would involve multiplying equations to cancel y², but that is rarely needed at A‑Level for intersections.
由线性方程解出 y = 7 − 2x，代入圆的方程：x² + (7 − 2x)² = 25。展开并化简，得到一个关于 x 的二次方程。这本质上是代入法；真正的消元法需要将方程相乘消去 y²，但在 A‑Level 的交点问题中很少需要这样做。
The key insight: any system consisting of a line and a conic (circle, parabola, ellipse, hyperbola) can be reduced to a quadratic equation in one variable by substitution. The discriminant of that quadratic then reveals how many times the line meets the curve.
关键点：任何由一条直线和一个圆锥曲线（圆、抛物线、椭圆、双曲线）组成的方程组，都可以通过代入化归为单变量的二次方程。该二次方程的判别式便能揭示直线与曲线相交的次数。
5. Intersection of a line and a quadratic curve: discriminant analysis | 直线与二次曲线相交：判别式分析
Ax² + Bx + C = 0. The discriminant Δ = B² − 4AC determines the nature of the intersection:
对于直线 y = mx + c 与二次曲线 y = ax² + bx + d，代入后得到 ax² + bx + d = mx + c → ax² + (b − m)x + (d − c) = 0。设方程为 Ax² + Bx + C = 0，判别式 Δ = B² − 4AC 决定了相交的性质：
Δ > 0: two distinct real roots → the line cuts the curve at two distinct points.
Δ > 0：两个相异实根 → 直线与曲线交于两个不同的点。
one repeated real root → the line is tangent to the curve (touches at exactly one point).
Δ = 0：一个重实根 → 直线与曲线相切（恰好接触于一点）。
Δ < 0: no real roots → the line does not intersect the curve.
Δ < 0：无实根 → 直线与曲线不相交。
This is extremely useful for problems that ask “Find the value of k for which the line y = 2x + k is tangent to the curve y = x² + 3x − 1.” You set the discriminant of the resulting quadratic to zero and solve for k.
这在解答“求使直线 y = 2x + k 与曲线 y = x² + 3x − 1 相切的 k 值”这类问题时极为有用。只需令所得二次方程的判别式为零，然后解出 k 即可。
Δ = 0 → tangent condition
Remember to always simplify the quadratic to standard form before calculating B² − 4AC. Misidentifying the coefficients is a common pitfall.
切记在计算 B² − 4AC 之前，要先将二次方程化为标准形式。错误地识别系数是一个常见陷阱。
6. Intersection of two quadratic curves | 两条二次曲线的交点
cles, substitution generally leads to a quartic (degree 4) equation. However, in Edexcel A‑ nce, the circle x² + y² = 4 and the parabola y² = 2x. Substituting y² = 2x into the circle gives x² + 2x = 4 → x² + 2x − 4 = 0, a simple quadratic. Always look for opportunities to replace y² or x² directly rather than expanding everything.
当两条曲线都是二次的，比如一个圆和一条抛物线，或者两个圆时，代入通常会导出一个四次（四次方）方程。然而，在 Edexcel A‑Level 的考题中，这些题目都经过精心设计，使得代入后往往可以提取公因式或降低方程的次数。例如，圆 x² + y² = 4 与抛物线 y² = 2x。将 y² = 2x 代入圆的方程，得到 x² + 2x = 4 → x² + 2x − 4 = 0，一个简单的二次方程。所以要善于寻找直接替换 y² 或 x² 的机会，而不是盲目地展开所有项。
If both equations are quadratic in x and y without an obvious substitution, subtracting one equation from the other can yield a linear relation between x and y, which can then be used for substitution. This is a powerful trick for intersecting circles or a circle and a parabola.
如果两个方程在 x 和 y 上都是二次的，又没有明显的代换方式，可以将两个方程相减，从而得到一个 x 与 y 的线性关系，再以此进行代入。这是解决两圆相交或圆与抛物线相交问题的一个强大技巧。
7. Points of intersection with parametric curves | 参数曲线的交点
so that g(t) = h(f(t)). This yields an equation solely in the parameter t. Solve for t, then obtain the coordinates (f(t), g(t)). Similarly, if two curves are both given parametrically, say C₁: (x₁(t), y₁(t)) and C₂: (x₂(s), y₂(s)), you set x₁(t) = x₂(s) and y₁(t) = y₂(s) and solve the simultaneous equations in t and s. These often rely on trigonometric identities when the parameter is an angle.
参数曲线由 x = f(t), y = g(t) 定义。要找到它与另一条曲线 C: y = h(x) 的交点，只需将参数表达式代入，得到 g(t) = h(f(t))，这样就形成了一个仅含参数 t 的方程。解出 t，然后求得坐标 (f(t), g(t))。类似地，如果两条曲线都用参数形式给出，例如 C₁: (x₁(t), y₁(t)) 和 C₂: (x₂(s), y₂(s))，则令 x₁(t) = x₂(s) 和 y₁(t) = y₂(s)，联立求解关于 t 和 s 的方程。当参数是角度时，这类问题经常需要运用三角恒等式。
Example: x = 2 cos t, y = sin t and x − 2y = 1
t + α) or squaring (with care for extraneous solutions) will yield t. Then x and y follow.
代入后得到 2 cos t − 2 sin t = 1 → cos t − sin t = ½。利用 R sin(t + α) 或两边平方（需注意增根）可求出 t，进而得到 x 和 y。
8. Multiple intersections and counting solutions | 多个交点与解的计数
A‑Level problems may ask for the number of intersection points without necessarily finding their exact coordinates. The discriminant method already showed how a line and a quadratic can have 0, 1, or 2 intersections. For higher degree polynomials, the fundamental theorem of algebra tells you there can be up to n intersections if a polynomial equation of degree n is involved. Graphical considerations and monotonic behaviour of functions can also help determine how many real solutions exist.
A‑Level 考题可能会要求你找出交点的个数，而不一定要计算出具体坐标。判别式法已经展示了直线与二次曲线可以有 0、1 或 2 个交点。对于更高次的多项式，代数基本定理告诉我们，如果涉及一个 n 次多项式方程，最多可以有 n 个交点。通过图像观察以及函数的单调性分析，也有助于判断存在多少个实数解。
For example, to determine how many times the curves y = eˣ and y = 2 − x intersect, you can define h(x) = eˣ −1 and h(1) = e − 1 > 0, there is at least one root. By checking the derivative h′(x) = eˣ + 1 > 0, h is strictly increasing, so there is exactly one intersection point.
例如，要判断曲线 y = eˣ 和 y = 2 − x 相交多少次，可以定义 h(x) = eˣ + x − 2 = 0。由于 h(0) = −1 且 h(1) = e − 1 > 0，至少存在一个根。再检查导数 h′(x) = eˣ + 1 > 0，说明 h 严格单调递增，因此恰好有一个交点。
9. Avoiding and identifying extraneous solutions | 避免与识别增根
When solving intersection equations, extraneous solutions can creep in, especially when squaring both sides, multiplying by an expression that could be zero, or manipulating trigonometric equations. Always substitute your final (x, y) pairs back into the original equations to verify they satisfy both. This is particularly important for parametric intersections, radical equations, and when you have divided by something that could be zero.
在求解交点方程时，增根可能会悄悄出现，尤其是在两边平方、乘以可能为零的表达式，或处理三角方程时。请务必将最终得到的 (x, y) 坐标代回原始方程，验证它们是否同时满足。这对于参数形式的交点、根式方程以及可能除以了零表达式的情形尤为重要。
A classic pitfall: solving √x + 3 yields x + 3 = (x − 3)² → x² − 7x + 6 = 0 → x = 1 or x = 6. Substituting back: x = 1 gives √4 = −2, false; x = 6 gives √9 raneous x = 1 emerged because squaring lost the sign condition (the right side must be non‑negative).
一个经典陷阱：解方程 √x + 3 = x − 3。平方后得 x + 3 = (x − 3)² → x² − 7x + 6 = 0 → x = 1 或 x = 6。代回原方程：x = 1 时 √4 = −2，不成立；x = 6 时 √9 = 3，成立。增根 x = 1 出现，是因为平方丢失了符号条件（右边必须非负）。
10. Geometric interpretation and tangency | 几何解释与相切
c solutions; they have clear geometric meaning. A tangent represents a double intersection— e concept of the derivative. For example, to prove that the line y = 4x − 4 is tangent to the parabola y = x², you can find their intersection by solving x² = 4x − 4 → x² − 4x + 4 = 0 → (x − 2)² = 0, giving a repeated root x = 2. The gradients at x = 2 are both 4, confirming tangency.
交点不单是代数解，它们有着明确的几何含义。切线代表一个二重交点——两个重合的点。在切点处，两条曲线（如果它们都可导）的梯度相等。这与微分和导数的概念相关联。例如，要证明直线 y = 4x − 4 与抛物线 y = x² 相切，可以通过求解 x² = 4x − 4 → x² − 4x + 4 = 0 → (x − 2)² = 0 得到重根 x = 2，从而找到它们的交点。在 x = 2 处，两者的梯度都是 4，进一步确认相切。
Similarly, two curves intersect orthogonally if, at the intersection point, the product of their gradients is −1. This can appear in synoptic questions linking coordinate geometry and calculus.
类似地，如果在交点处两条曲线的梯度乘积为 −1，则它们正交相交。这种问题可能出现在综合坐标几何与微积分的考题中。
11. Worked example: thorough walkthrough | 典型例题：详细讲解
Problem: x² + y² − 4x + 2y − 20 = 0. Determine also the distance between these intersection points.
问题：求直线 y = 2x − 1 与圆 x² + y² − 4x + 2y − 20 = 0 的交点坐标，并计算这两个交点之间的距离。
ectly: (2x − 1)² = 4x² − 4x + 1; 2y = 2(2x − 1) = 4x − 2. So total: x² + 4x² − 4x + 1 − 4x + 4x − 2 − 20 = 5x² − 4x − 21 = 0. Yes.
代入 y：x² + (2x − 1)² − 4x + 2(2x − 1) − 20 = 0。展开：x² + 4x² − 4x + 1 − 4x + 4x − 2 − 20 = 0 → 5x² − 4x − 21 = 0。（注意 −4x + 4x 抵消，需仔细。）正确化简：(2x − 1)² = 4x² − 4x + 1；2y = 2(2x − 1) = 4x − 2。总和为 x² + 4x² − 4x + 1 − 4x + 4x − 2 − 20 = 5x² − 4x − 21 = 0。没错。
= (−1 ± 2√109)/5. The intersection points are P = ((2 + √109)/5, (−1 + 2√109)/5) and Q = ((2 − √109)/5, (−1 − 2√109)/5). Distance PQ: √[(Δx)² + (Δy)²] = √[ ( (2√109)/5 )² + ( (4√109)/5 )² ] = √[ (4×109)/25 + (16×109)/25 ] = √[20×109/25] = √[(4×109)/5] = (2√109)/√5? Actually 20/25 = 4/5, so distance = √4×109/5 = 2√109/5 = (2√545)/5 after rationalising. But exact form √436/5 = √436/√5 = 2√109/√5 = (2√545)/5. Keep surd form.
求解：判别式 Δ = (−4)² − 4(5)(−21) = 16 + 420 = 436 = 4 × 109。因此 x = [4 ± 2√109] / 10 = [2 ± √109] / 5。计算 y：y = 2x − 1 = (4 ± 2√109)/5 − 1 = (4 ± 2√109 − 5)/5 = (−1 ± 2√109)/5。交点为 P = ((2 + √109)/5, (−1 + 2√109)/5) 和 Q = ((2 − √109)/5, (−1 − 2√109)/5)。距离 /5 )² + ( (4√109)/5 )² ] = √[ (4×109)/25 + (16×109)/25 ] = √[20×109/25] = √[(4×109)/5] = 2√109/5 = (2√545)/5（有理化后）。保留根式形式即可。
12. Common mistakes and final tips | 常见错误与最后提示
Mistakes often arise from:
- Forgetting to substitute back to find the other coordinate.
- Mishandling signs when expanding brackets, especially with subtraction.
ing terms, resulting in missing solutions (e.g., dividing by x without considering x = 0).
- Failing to check the domain or algebraic restrictions, leading to extraneous roots.
- In parametric problems, not verifying that the parameter values produce coordinates on both curves, or assuming a single t works for both without checking the other curve’s parameterisation.
常见错误来源：
- 忘记代回求另一坐标。
- 展开括号时符号处理不当，尤其是带有减号时。
- 约项时丢失因式，造成漏解（例如，没有考虑 x = 0 的情况就除以 x）。
- 未检查定义域或代数限制，导致增根。
- 在参数问题中，未验证参数值是否使坐标同时位于两条曲线上，或未检查另一条曲线的参数化就假设单一的 t 可行。
Always draw a quick sketch if possible. A rough graph can reveal how many intersections to expect and whether your algebraic answers are plausible. When dealing with tangency, remember that the discriminant is your best friend. And finally, practise with past Edexcel questions: the more you expose yourself to varied function types—polynomials, trigonometric, exponential, and parametric—the more confident you will be.
只要可能，随手画一个草图。粗略的图像能让你预判交点的个数，并检验代数解是否合理。处理相切问题时，请记住判别式是你最好的帮手。最后，多练习 Edexcel 历年真题：接触的函数类型越丰富——多项式、三角函数、指数函数以及参数方程——你考场上就会越自信。
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