📚 Points of Intersection | 曲线的交点
In Edexcel A Level Mathematics, finding points of intersection is a core skill that appears in coordinate geometry, quadratics, circles and parametric equations. A point of intersection is simply a point that lies on two graphs at the same time, so its coordinates satisfy both equations simultaneously. This article explains the algebraic methods, discriminant conditions and exam techniques you need to solve intersection problems accurately.
在 Edexcel A Level 数学中，求交点是坐标几何、二次函数、圆和参数方程中的核心技能。交点就是同时位于两个图形上的点，因此它的坐标同时满足两个方程。本文讲解求解交点问题所需的代数方法、判别式条件和考试技巧。
1. The Core Idea: Same Coordinates, Same Point | 核心思想：同一坐标，同一点
If two curves y = f(x) and y = g(x) intersect at a point P, then P has coordinates (x, y) that satisfy both equations simultaneously. In algebraic terms, finding intersections is the same as solving simultaneous equations.
如果两条曲线 y = f(x) 与 y = g(x) 在点 P 处相交，那么 P 的坐标 (x, y) 同时满足两个方程。从代数角度看，求交点等价于解联立方程组。
For a point to lie on both curves, its x-coordinate must make f(x) equal to g(x). The y-coordinate is then found by substituting the x-value into either original equation. Always write intersection points as coordinate pairs, not just x-values.
一个点要同时在两条曲线上，其 x 坐标必须使 f(x) = g(x)。然后将 x 值代入任意一个原方程求出 y 坐标。交点要写成坐标对，不能只写 x 值。
2. Linear and Quadratic Intersections | 直线与二次曲线的交点
The most common Edexcel problem involves a straight line y = mx + c and a quadratic curve y = ax² + bx + c. Substituting the line into the quadratic gives a single quadratic equation in x, which can then be solved by factorising or using the quadratic formula.
Edexcel 考试中最常见的问题是直线 y = mx + c 与二次曲线 y = ax² + bx + c 的交点。将直线代入二次曲线，可得到关于 x 的一元二次方程，然后用因式分解或求根公式求解。
For example, to find where y = x + 1 meets y = x² − 2x − 3, write x + 1 = x² − 2x − 3. Rearranging gives x² − 3x − 4 = 0, so x = 4 or x = −1. Substituting back gives the points (4, 5) and (−1, 0).
例如，求 y = x + 1 与 y = x² − 2x − 3 的交点：令 x + 1 = x² − 2x − 3，整理得 x² − 3x − 4 = 0，因此 x = 4 或 x = −1。代回原方程得交点为 (4, 5) 和 (−1, 0)。
3. Using the Discriminant | 利用判别式
After substitution, the number of intersections can be determined from the discriminant of the resulting quadratic ax² + bx + c = 0. The discriminant is Δ = b² − 4ac.
代入后，所得一元二次方程 ax² + bx + c = 0 的判别式可以判断交点的个数。判别式为 Δ = b² − 4ac。
| Discriminant Δ | Real roots | Intersections |
|---|---|---|
| Δ > 0 | Two distinct real roots | Two distinct points |
| Δ = 0 | One repeated real root | Tangent, one point |
| Δ < 0 | No real roots | No intersection |
If Δ > 0, the quadratic has two distinct real roots, so there are two distinct points of intersection. If Δ = 0, there is one repeated root, which corresponds to the line touching the curve at a single point. If Δ < 0, there are no real roots, so the line does not meet the curve.
若 Δ > 0，方程有两个不同实根，因此直线与曲线有两个不同交点；若 Δ = 0，方程有一个重根，对应直线与曲线在一点处相切；若 Δ < 0，方程没有实根，直线与曲线不相交。
4. Tangents as Repeated Roots | 切线与重根
When a line is tangent to a curve, the point of tangency is a double intersection. Algebraically, the equation formed by substitution has a repeated root, so the discriminant is exactly zero.
当直线与曲线相切时，切点是一个二重交点。从代数上看，代入后得到的方程有重根，因此判别式恰好为零。
This idea is used to find unknown parameters. For example, if y = 2x + k is tangent to y = x² + 3x + 1, setting 2x + k = x² + 3x + 1 gives x² + x + 1 − k = 0. For tangency, Δ = 1² − 4(1)(1 − k) = 0, so k = 3/4.
这一思想常用于求未知参数。例如，若 y = 2x + k 与 y = x² + 3x + 1 相切，令 2x + k = x² + 3x + 1，得 x² + x + 1 − k = 0。相切条件为 Δ = 1² − 4(1)(1 − k) = 0，解得 k = 3/4。
5. Intersections of Two Curves | 两条曲线的交点
For two general curves y = f(x) and y = g(x), equate the two expressions for y and solve f(x) = g(x). This gives the x-coordinates of all intersection points.
对于一般曲线 y = f(x) 和 y = g(x)，令两个 y 的表达式相等，解方程 f(x) = g(x)，即可得到所有交点的 x 坐标。
After finding each x-value, substitute it back into either original equation to find the corresponding y-value. Do not use a combined expression that may not represent the original curve, because rearranging can sometimes introduce errors.
求出每个 x 值后，将其代回任意一个原方程，得到对应的 y 值。不要使用合并后的表达式，因为它可能不能代表原曲线，移项时也容易出错。
If the equation f(x) = g(x) is not linear or quadratic, use factorisation, the quadratic formula, or graphical or numerical methods as appropriate. In Edexcel exams, the resulting equation is usually designed to be solvable by standard methods.
如果方程 f(x) = g(x) 不是一次或二次方程，应根据情况使用因式分解、求根公式或图形/数值方法。在 Edexcel 考试中，所得方程通常都可以用标准方法求解。
6. Circles and Lines | 圆与直线的交点
Edexcel coordinate geometry often asks for intersections between a line and a circle. A circle with centre (a, b) and radius r has equation (x − a)² + (y − b)² = r².
Edexcel 坐标几何经常考查直线与圆的交点。圆心为 (a, b)、半径为 r 的圆方程为 (x − a)² + (y − b)² = r²。
To find intersections, rearrange the line into the form y = mx + c or x = d, substitute into the circle equation, and solve the resulting quadratic in one variable. Then find the other coordinate for each solution.
求交点时，先将直线写成 y = mx + c 或 x = d 的形式，代入圆方程，再解关于一个变量的一元二次方程。然后对每个解求出另一个坐标。
Alternatively, compare the perpendicular distance from the centre to the line with the radius. If the distance is less than r, the line cuts the circle at two points; if equal to r, the line is tangent; if greater than r, there is no intersection.
另一种方法是比较圆心到直线的垂直距离与半径。若距离小于 r，直线与圆相交于两点；若等于 r，直线与圆相切；若大于 r，则没有交点。
7. Parametric Curves | 参数曲线的交点
If a curve is given parametrically by x = p(t), y = q(t), intersections with another curve can be found by converting to Cartesian form or by substituting parametric expressions into the other equation.
如果曲线由参数方程 x = p
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