Points of Intersection | 交点

📚 Points of Intersection | 交点

When you sketch two curves on the same coordinate plane, the places where they cross are called points of intersection. In A-Level Mathematics, finding these points is a core skill that links algebra and geometry: you need to solve two equations simultaneously. The coordinates of any intersection point must satisfy both equations, so the problem reduces to solving a system of equations. Understanding how many intersections exist and what they represent geometrically is essential for topics ranging from straight lines and circles to parametric curves.

当你在同一坐标系中绘制两条曲线时，它们交叉的地方就称为交点。在 A-Level 数学中，求交点是一项连接代数与几何的核心技能：你需要同时求解两个方程。交点的坐标必须同时满足这两个方程，因此问题就转化为求解方程组。理解交点的数量及其几何意义，对于从直线、圆到参数曲线等众多主题都是至关重要的。

1. Setting Up the Simultaneous Equations | 联立方程的建立

Suppose we have two curves defined by equations f(x, y) = 0 and g(x, y) = 0. To find their intersection points, we write them as a system: f(x, y) = 0 and g(x, y) = 0. The solutions (x, y) that satisfy both are the intersection points. In many Edexcel exam questions, one curve is a straight line and the other is a quadratic curve such as a circle, parabola, or ellipse.

假设我们有由方程 f(x, y) = 0 和 g(x, y) = 0 定义的两条曲线。为了找到它们的交点，我们将其写为一个方程组：f(x, y) = 0 和 g(x, y) = 0。同时满足这两个方程的 (x, y) 解就是交点。在许多 Edexcel 考题中，一条曲线是直线，另一条是二次曲线，例如圆、抛物线或椭圆。

The most effective strategy is to rearrange one equation to express one variable in terms of the other, then substitute it into the second equation. This elimination process usually leaves you with a single-variable quadratic equation, which can then be solved using factorisation, completing the square, or the quadratic formula.

最有效的策略是将其中一个方程变形，用一个变量表示另一个变量，然后将其代入第二个方程。这个消元过程通常会得到一个一元二次方程，然后可以通过因式分解、配方法或求根公式来求解。

2. The Substitution Method in Detail | 代入法详解

Take a line with equation y = mx + c and a circle with equation x² + y² = r². Substitute the expression for y from the line into the circle’s equation: x² + (mx + c)² = r². Expand and simplify to obtain a quadratic in x: (1 + m²)x² + 2mcx + (c² − r²) = 0. Solving this gives the x-coordinates of any intersection points. Once x is known, substitute back into y = mx + c to find the corresponding y-coordinates.

以直线 y = mx + c 和圆 x² + y² = r² 为例。将直线中 y 的表达式代入圆的方程：x² + (mx + c)² = r²。展开并化简后可得到关于 x 的二次方程：(1 + m²)x² + 2mcx + (c² − r²) = 0。解此方程可得到交点的 x 坐标。求出 x 后，再代回 y = mx + c 即可得到相应的 y 坐标。

If the second curve is given in a different form, say y = ax² + bx + c, the process is similar. Substitute the linear expression for y into the quadratic and collect like terms. The resulting equation will still be quadratic in x, making further analysis straightforward using the discriminant.

如果第二条曲线是其他形式，比如 y = ax² + bx + c，过程类似。将直线的 y 表达式代入二次曲线，并合并同类项。最终得到关于 x 的方程仍然是二次的，便于利用判别式进行进一步分析。

3. The Role of the Discriminant | 判别式的作用

For a quadratic equation ax² + bx + c = 0 (with a ≠ 0), the discriminant Δ is given by Δ = b² − 4ac. The value of Δ determines the nature of the roots, and therefore the number of intersection points between the two original curves:

对于二次方程 ax² + bx + c = 0（其中 a ≠ 0），判别式 Δ 定义为 Δ = b² − 4ac。Δ 的值决定了根的性质，因此也决定了原两条曲线交点的个数：

• Δ > 0: Two distinct real roots → two distinct intersection points.
  Δ > 0：有两个不相等的实根 → 两个不同的交点。
• Δ = 0: One repeated real root → the line touches the curve at exactly one point (it is a tangent).
  Δ = 0：有一个重根 → 直线恰好在一个点处接触曲线（即相切）。
• Δ < 0: No real roots → the curves do not meet at all.
  Δ < 0：没有实根 → 两条曲线永不相交。

In many Edexcel questions, you are not always required to fully solve for the coordinates; you may only need to use the discriminant to show that a line is tangent to a curve, or to find conditions on a parameter for which there are two, one, or zero points of intersection.

在许多 Edexcel 考题中，你并不总是需要完全求出坐标；你可能只需要利用判别式来证明一条直线与一条曲线相切，或者找出使交点为两个、一个或零个的参数条件。

4. Tangents and Repeated Roots | 切线与重根

When Δ = 0, the quadratic has equal roots, and the line is a tangent to the curve. Geometrically, this means the line just touches the curve without crossing it. For example, the line y = mx + c is tangent to the circle x² + y² = r² if the perpendicular distance from the centre to the line equals the radius, which algebraically corresponds to Δ = 0.

当 Δ = 0 时，二次方程有等根，直线就是曲线的切线。从几何角度看，这意味着直线刚好接触曲线而不穿过它。例如，直线 y = mx + c 与圆 x² + y² = r² 相切的条件是圆心到直线的垂直距离等于半径，代数上这正对应于 Δ = 0。

In pure algebraic terms, if the final quadratic in x is (px + q)² = 0, then x = −q/p is the sole x-coordinate of the point of contact. Substituting back gives the y-coordinate, revealing the point of tangency. Exam questions often ask you to find the value of a constant (such as c or m) that makes a line tangent to a given curve, and setting Δ = 0 is the standard method.

用纯代数语言来说，如果最终关于 x 的二次方程为 (px + q)² = 0，那么 x = −q/p 就是切点的唯一 x 坐标。代回可求出 y 坐标，从而得到切点。考题经常要求你找出使直线成为给定曲线切线的常数（如 c 或 m），而设 Δ = 0 是标准方法。

5. Intersection of a Line and a Parabola | 直线与抛物线的交点

A typical problem combines y = mx + c with y = ax² + bx + c_curve. After substitution, you obtain ax² + bx + c_curve = mx + c_line, which rearranges to ax² + (b − m)x + (c_curve − c_line) = 0. The discriminant of this quadratic tells you how many times the line cuts the parabola. Remember to label coefficients carefully to avoid confusion between the two ‘c’s.

一个典型的问题是将 y = mx + c 与 y = ax² + bx + c_曲线 联立。代入后得到 ax² + bx + c_曲线 = mx + c_直线，整理后为 ax² + (b − m)x + (c_曲线 − c_直线) = 0。这个二次方程的判别式就能告诉你直线与抛物线相交的次数。注意仔细标记系数，避免两个 ‘c’ 混淆。

Once you have solved for x, always plug the values back into the linear equation to find y; this is usually simpler than using the quadratic equation for y. Also check that any restrictions on the domain (e.g. from a square root) are satisfied by your solutions.

一旦求出 x，一定要把值代回线性方程求 y；这通常比用二次方程求 y 更简单。同时，检查你的解是否满足定义域的任何限制（例如来自平方根的限制）。

6. Intersection of Two Quadratic Curves | 两条二次曲线的交点

When both curves are quadratic, such as a circle and a parabola, or two parabolas, the algebra can lead to a quartic (degree 4) equation. However, in Edexcel questions, the curves are often chosen so that elimination reduces the system to a quadratic. For instance, equating two expressions for y² or using substitution cleverly can keep the degree low.

当两条曲线都是二次曲线时，例如圆和抛物线，或者两条抛物线，代数上可能会导致一个四次方程。不过，在 Edexcel 的题目中，曲线通常会经过精心选择，使得消元后得到一个二次方程。例如，令两个 y² 的表达式相等，或巧妙地使用代入法，都可以降低方程的次数。

Consider the curves y² = 4x and x² + y² = 9. Substituting y² = 4x into the circle equation gives x² + 4x − 9 = 0, which is a quadratic in x. Solving yields two x-values, and then y = ±√(4x) gives up to four intersection points. Always consider both positive and negative square roots where applicable.

考虑曲线 y² = 4x 和 x² + y² = 9。将 y² = 4x 代入圆的方程，得到 x² + 4x − 9 = 0，这是一个关于 x 的二次方程。求解得到两个 x 值，然后由 y = ±√(4x) 可得最多四个交点。在有平方根的情况下，务必同时考虑正负平方根。

7. Parametric Equations and Intersection | 参数方程与交点

Some curves are defined parametrically: x = f(t), y = g(t). To find their intersection with another curve, say a line y = mx + c, you substitute the parametric expressions directly: g(t) = m f(t) + c. This gives an equation in the parameter t. Solve for t, then substitute back into f(t) and g(t) to obtain the (x, y) coordinates of intersection.

有些曲线是用参数方程定义的：x = f(t), y = g(t)。要找到它们与另一条曲线（如直线 y = mx + c）的交点，你需要直接代入参数表达式：g(t) = m f(t) + c。这样会得到一个关于参数 t 的方程。解出 t，然后代回 f(t) 和 g(t) 即可得到交点的 (x, y) 坐标。

Be careful with the range of t if it is given. Sometimes a parametric curve only covers part of the curve, so only those t-values within the specified interval correspond to actual points on the curve. Also remember that different t-values might produce the same (x, y) point, so checking for uniqueness may be needed.

如果给出了 t 的取值范围，要多加注意。有时参数曲线只覆盖曲线的一部分，因此只有在指定区间内的 t 值才对应曲线上的实际点。此外，不同的 t 值可能产生相同的 (x, y) 点，所以有时需要验证交点的唯一性。

8. Geometric Interpretation and Graph Sketching | 几何意义与图形草图

Before diving into algebra, it is wise to sketch the two curves. A rough graph can reveal the approximate number of intersection points and alerts you to potential extraneous solutions. For example, a line can intersect a circle in 0, 1, or 2 points; a sketch immediately suggests which discriminant condition to expect.

在深入代数计算之前，明智的做法是画出两条曲线的草图。一个粗略的图形能揭示交点的大致数目，并提醒你注意可能的增根。例如，一条直线与一个圆可能交于 0 个、1 个或 2 个点；草图能立刻让人预期何种判别式条件。

Understanding the geometric meaning of the discriminant helps you form a mental picture. A positive discriminant corresponds to two crossing points, zero to a tangent, and negative to no intersection. This geometric perspective can be particularly helpful when solving problems involving inequalities: for a given m, does the line lie entirely above a parabola?

理解判别式的几何意义有助于形成直观印象。判别式为正对应两个交点，为零对应相切，为负则没有交点。这种几何视角在解决涉及不等式的问题时尤其有用：对于给定的 m，直线是否整个位于抛物线的上方？

9. Worked Example: Line and Circle | 例题：直线与圆

Problem: Find the coordinates of the points where the line y = 2x + 1 intersects the circle x² + y² = 25.

题目：求直线 y = 2x + 1 与圆 x² + y² = 25 的交点坐标。

Step 1: Substitute y = 2x + 1 into x² + y² = 25.
步骤 1：将 y = 2x + 1 代入 x² + y² = 25。

This gives x² + (2x + 1)² = 25. Expand: x² + 4x² + 4x + 1 = 25 → 5x² + 4x − 24 = 0.
得到 x² + (2x + 1)² = 25。展开：x² + 4x² + 4x + 1 = 25 → 5x² + 4x − 24 = 0。

Step 2: Solve the quadratic 5x² + 4x − 24 = 0. Use the quadratic formula: x = [−4 ± √(16 + 480)] / 10 = [−4 ± √496] / 10. √496 = 4√31, so x = [−4 ± 4√31] / 10 = [−2 ± 2√31] / 5.
步骤 2：解二次方程 5x² + 4x − 24 = 0。使用求根公式：x = [−4 ± √(16 + 480)] / 10 = [−4 ± √496] / 10。√496 = 4√31，因此 x = [−4 ± 4√31] / 10 = [−2 ± 2√31] / 5。

Step 3: Find corresponding y-values using y = 2x + 1. For x = (−2 + 2√31)/5, y = 2((−2 + 2√31)/5) + 1 = (−4 + 4√31)/5 + 5/5 = (1 + 4√31)/5. For the other x, y = (1 − 4√31)/5.
步骤 3：利用 y = 2x + 1 求出相应的 y 值。对于 x = (−2 + 2√31)/5，y = 2((−2 + 2√31)/5) + 1 = (−4 + 4√31)/5 + 5/5 = (1 + 4√31)/5。对于另一个 x，y = (1 − 4√31)/5。

The two intersection points are therefore ( (−2 + 2√31)/5 , (1 + 4√31)/5 ) and ( (−2 − 2√31)/5 , (1 − 4√31)/5 ). Always present the coordinates as ordered pairs.

因此，两个交点为 ( (−2 + 2√31)/5 , (1 + 4√31)/5 ) 和 ( (−2 − 2√31)/5 , (1 − 4√31)/5 )。要始终将坐标写成有序对的形式。

10. Worked Example: Two Parabolas | 例题：两条抛物线

Problem: Determine the points where y = x² − 2x − 3 meets y = −x² + 4.
题目：求曲线 y = x² − 2x − 3 与 y = −x² + 4 的交点。

Step 1: Set the two expressions for y equal: x² − 2x − 3 = −x² + 4.
步骤 1：令两个 y 的表达式相等：x² − 2x − 3 = −x² + 4。

Step 2: Rearrange to a standard quadratic: 2x² − 2x − 7 = 0 → x² − x − 3.5 = 0, or multiply by 2: 2x² − 2x − 7 = 0.
步骤 2：整理为标准二次方程：2x² − 2x − 7 = 0 → x² − x − 3.5 = 0，或者保持 2x² − 2x − 7 = 0。

Step 3: Solve using the quadratic formula: x = [2 ± √(4 + 56)] / 4 = [2 ± √60] / 4 = [2 ± 2√15] / 4 = [1 ± √15] / 2.
步骤 3：使用求根公式求解：x = [2 ± √(4 + 56)] / 4 = [2 ± √60] / 4 = [2 ± 2√15] / 4 = [1 ± √15] / 2。

Step 4: Find y by substituting into either original equation, e.g. y = −x² + 4. For x = (1 + √15)/2, x² = (1 + 2√15 + 15)/4 = (16 + 2√15)/4 = 4 + (√15)/2, so y = −[4 + (√15)/2] + 4 = −(√15)/2. The other point has y = (√15)/2.
步骤 4：代入任一原方程求 y，例如 y = −x² + 4。对于 x = (1 + √15)/2，x² = (1 + 2√15 + 15)/4 = (16 + 2√15)/4 = 4 + (√15)/2，因此 y = −[4 + (√15)/2] + 4 = −(√15)/2。另一点有 y = (√15)/2。

The intersection points are ( (1 + √15)/2 , −√15/2 ) and ( (1 − √15)/2 , √15/2 ). Double-check by plugging into the first parabola equation.

交点坐标为 ( (1 + √15)/2 , −√15/2 ) 和 ( (1 − √15)/2 , √15/2 )。代回第一条抛物线方程进行复核。

11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Mistake 1: Forgetting to find both coordinates. After obtaining x, students often stop. Always substitute back to find the full (x, y) pair. Mistake 2: Algebraic slips when expanding (mx + c)². Remember (a + b)² = a² + 2ab + b². Mistake 3: Ignoring the negative square root when solving y² = expression. This can cost you half of the intersection points. Mistake 4: Misidentifying a, b, c in the quadratic when terms are rearranged. Write the equation in the form ax² + bx + c = 0 before applying the discriminant.

错误 1：忘记求两个坐标。求出 x 后，学生往往就停下了。一定要代回求出完整的 (x, y) 坐标。错误 2：展开 (mx + c)² 时出现代数错误。记住 (a + b)² = a² + 2ab + b²。错误 3：在解 y² = 表达式时忽略了负平方根。这可能导致你遗漏一半的交点。错误 4：在整理各项时错误识别二次方程中的 a、b、c。在应用判别式之前，务必将方程写成 ax² + bx + c = 0 的形式。

Mistake 5: Not considering domain restrictions, especially with square root functions or parametric ranges. Mistake 6: Using the quadratic formula with an incorrect denominator (e.g. 2a but a is the coefficient of x²). Double-check your a, b, c. Finally, many marks are lost by writing coordinates in the wrong order. Always write (x, y).

错误 5：未考虑定义域限制，尤其是存在平方根函数或参数范围时。错误 6：使用求根公式时分母错误（例如 2a 而 a 是 x² 的系数）。再次检查 a、b、c。最后，很多分数因坐标书写顺序错误而丢失。始终写成 (x, y) 的形式。

12. Summary and Key Takeaways | 总结与要点

Points of intersection are found by solving the equations of the curves simultaneously. The usual method is substitution, leading to a single-variable equation — often a quadratic. The discriminant of that quadratic tells you immediately how many intersection points exist: two (>0), one tangent (=0), or none (<0). Always complete the solution by finding both coordinates and present them as ordered pairs.

交点的求法是对曲线的方程进行联立求解。通常的方法是代入法，从而得出一个单变量方程——通常是二次方程。该二次方程的判别式能立刻告诉你交点的个数：两个（>0）、一个相切（=0）或没有（<0）。一定要通过求出两个坐标来完善解答，并以有序对的形式呈现。

Work carefully through the algebra, paying attention to signs and domains, and you will master this topic. Points of intersection questions are common across pure mathematics papers, and the concepts extend to differentiation, integration, and even mechanics when interpreting graphs of motion.

在代数推导中要仔细，注意符号和定义域，你就能掌握这个主题。交点问题在纯数试卷中十分常见，其概念还可延伸到微分、积分，甚至在解释运动图像时也会用到。

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