# 进阶坐标系

> 爱德思国际A-Level进阶数学 · IAL 2018版 FP3
> 来源: https://www.owlsprep.com/zh/study/edexcel-ial-further-math-fp3-further-coordinate-systems/

本指南覆盖爱德思IAL FP3进阶坐标系的全部内容，包括椭圆/双曲线方程、离心率、焦点-准线性质、切线/法线以及简单轨迹问题，完全贴合2018年WFM03考纲要求。

**先修:** [FP1坐标系（抛物线、等轴双曲线）](https://www.owlsprep.com/zh/study/edexcel-ial-further-math-fp1-coordinate-systems/); [隐函数微分与参数微分](https://www.owlsprep.com/zh/study/edexcel-ial-math-a2-differentiation/)

## 学习目标

- 推导并使用椭圆与双曲线的直角坐标方程和参数方程
- 应用椭圆与双曲线的焦点-准线性质以及离心率公式
- 求解椭圆与双曲线的切线和法线方程，并牢记y=mx+c形式的相切条件
- 求解涉及上述圆锥曲线的简单轨迹问题

## 椭圆与双曲线的标准方程

FP3考察的两类新圆锥曲线是椭圆和普通双曲线，所有考试题目中二者的长轴均为水平方向、与x轴对齐。

**标准椭圆** — 其直角坐标方程为$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$，其中$a > b$，参数方程为$x = a \cos t$、$y = b \sin t$，可通过三角恒等式$\cos^2 t + \sin^2 t = 1$推导得到。

**标准双曲线** — 其直角坐标方程为$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$，有两种可用的参数形式：$x = a \sec t$、$y = b \tan t$（利用恒等式$\\text{sec}^2 t - \tan^2 t = 1$，覆盖双曲线的两支），或$x = a \cosh t$、$y = b \sinh t$（利用恒等式$\\text{cosh}^2 t - \sinh^2 t = 1$，仅覆盖$x \geq a$的右支）。

**例题:** 已知圆锥曲线的参数方程为$x = 3 \sec t$、$y = 5 \tan t$，求其直角坐标方程，并说明该曲线是椭圆还是双曲线。

1. 整理参数方程，分离出三角函数：$\frac{x}{3} = \sec t$，$\frac{y}{5} = \tan t$
2. 将上述结果代入三角恒等式$\text{sec}^2 t - \tan^2 t = 1$：
3. $$\left(\frac{x}{3}\right)^2 - \left(\frac{y}{5}\right)^2 = 1$$
4. 化简得到直角坐标方程：$\frac{x^2}{9} - \frac{y^2}{25} = 1$，该曲线为双曲线。

> **考试提示:** 对于限定在右支（$x>0$）的双曲线问题，建议使用双曲函数参数形式，以避免正割函数取负值带来的问题。

*计算器:* allowed

## 离心率、焦点与准线

所有圆锥曲线都可以通过决定其形状的离心率$e$定义。$a$、$b$和$e$的关系式，以及焦点坐标和准线方程都在公式手册中给出，但你必须能够正确应用这些公式。

| Conic | Eccentricity range | $b^2$ relation to $a,e$ | Foci coordinates | Directrix equations | Asymptotes |
| --- | --- | --- | --- | --- | --- |
| Ellipse | $e < 1$ | $b^2 = a^2(1-e^2)$ | $(\pm ae, 0)$ | $x = \pm \frac{a}{e}$ | None |
| Hyperbola | $e > 1$ | $b^2 = a^2(e^2 - 1)$ | $(\pm ae, 0)$ | $x = \pm \frac{a}{e}$ | $y = \pm \frac{b}{a}x$ |

**例题:** 已知椭圆方程为$\frac{x^2}{25} + \frac{y^2}{9} = 1$，计算其离心率、焦点坐标和准线方程。

1. 从标准方程中确定$a$和$b$的值：$a^2 = 25 \implies a=5$，$b^2=9 \implies b=3$
2. 代入椭圆的离心率公式：$9 = 25(1 - e^2)$
3. $$1 - e^2 = \frac{9}{25} \implies e^2 = \frac{16}{25} \implies e = \frac{4}{5}$$
4. 计算焦点坐标：$\pm ae = \pm 5 \times \frac{4}{5} = \pm 4$，因此焦点为$(4,0)$和$(-4,0)$
5. 计算准线方程：$\pm \frac{a}{e} = \pm \frac{5}{4/5} = \pm \frac{25}{4}$，因此准线为$x = 6.25$和$x = -6.25$

> **考试提示:** 对于椭圆，$e < 1$，因此$ae < a$（焦点位于椭圆内部）且$a/e > a$（准线位于椭圆外部），你可以利用这一性质快速核验答案是否合理。

*计算器:* allowed

## 圆锥曲线的切线与法线

你可以通过隐函数求导（适用于直角坐标方程）或参数求导（适用于参数坐标）来求解切线和法线的斜率。直线$y = mx + c$与圆锥曲线相切的条件必须牢记，该条件不会在公式手册中给出。

- Ellipse tangency condition: $c^2 = a^2 m^2 + b^2$
- Hyperbola tangency condition: $c^2 = a^2 m^2 - b^2$

**例题:** 求双曲线$\frac{x^2}{16} - \frac{y^2}{9} = 1$在$t = \frac{\pi}{4}$处的切线方程（使用参数形式$x = 4 \sec t$，$y = 3 \tan t$）。

1. 求$t = \frac{\pi}{4}$处的点坐标：$\text{sec}(\pi/4) = \sqrt{2}$，$\tan(\pi/4) = 1$，因此$x = 4\sqrt{2}$，$y=3$
2. 通过参数求导计算斜率：$\frac{dx}{dt} = 4 \sec t \tan t$，$\frac{dy}{dt} = 3 \text{sec}^2 t$
3. $$\frac{dy}{dx} = \frac{3 \text{sec}^2 t}{4 \sec t \tan t} = \frac{3 \sec t}{4 \tan t} = \frac{3}{4 \sin t}$$
4. 代入$t = \frac{\pi}{4}$：$\frac{dy}{dx} = \frac{3}{4 \times (\sqrt{2}/2)} = \frac{3}{2\sqrt{2}}$
5. 使用点斜式写出切线方程：$y - 3 = \frac{3}{2\sqrt{2}}(x - 4\sqrt{2})$
6. 化简为标准形式：$3x - 2\sqrt{2}y = 6\sqrt{2}$

> **考试提示:** 请务必使用相切条件验证你的切线方程，以排查计算错误：以上述示例为例，$c = -3$，$c^2 =9$，且$a^2 m^2 - b^2 = 16 \times (9/8) -9 = 9$，二者结果完全吻合。

*计算器:* allowed

## 简单轨迹问题

轨迹是满足给定条件的所有点的集合（例如切线的中点、两条法线的交点）。绝大多数FP3轨迹问题最简便的解法是：先使用圆锥曲线上点的参数坐标，将轨迹点的坐标用参数表示，再消去参数得到直角坐标方程。

**例题:** 求椭圆$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$的焦点$(ae, 0)$与椭圆上任意一点$P$连线所得线段的中点轨迹。

1. 设点$P$的参数坐标为$(a \cos t, b \sin t)$，中点为$M(h,k)$
2. 用参数$t$表示中点坐标：$h = \frac{a \cos t + ae}{2}$，$k = \frac{b \sin t + 0}{2}$
3. 整理方程以分离三角函数：$\frac{2h - ae}{a} = \cos t$，$\frac{2k}{b} = \sin t$
4. 使用三角恒等式$\cos^2 t + \sin^2 t = 1$消去参数$t$：
5. $$\left(\frac{2h - ae}{a}\right)^2 + \left(\frac{2k}{b}\right)^2 = 1$$
6. 将$h$替换为$x$、$k$替换为$y$，得到最终的轨迹方程：$\frac{(2x - ae)^2}{a^2} + \frac{4y^2}{b^2} = 1$，该轨迹为另一个椭圆。

> **考试提示:** 在处理轨迹问题时，只要条件允许请优先使用参数坐标，与直接使用直角坐标相比，参数坐标可以减少你需要处理的变量数量。

*计算器:* allowed

## 常见错误

- **错误做法:** Using the ellipse eccentricity formula $b^2 = a^2(1-e^2)$ for hyperbolas
  - 原因: Hyperbolas have $e>1$, so this would give a negative value for $b^2$, which is impossible
  - 正确做法: Remember the sign flips for hyperbolas: $b^2 = a^2(e^2 - 1)$
- **错误做法:** Using the ellipse tangency condition $c^2 = a^2m^2 + b^2$ for hyperbolas
  - 原因: The hyperbola tangency condition has a minus sign, so using the wrong sign will give an invalid value for $c$
  - 正确做法: Memorize the difference: ellipse uses +, hyperbola uses - for the tangency condition
- **错误做法:** Swapping foci and directrix formulae: using $x = \pm ae$ for directrices and $\pm a/e$ for foci
  - 原因: This gives coordinates/equations that are too large or small, leading to lost method marks
  - 正确做法: Sanity check with ellipse $e<1$: $ae < a$ so foci are inside the ellipse, $a/e > a$ so directrices are outside
- **错误做法:** Only using the sec-tan parametric form for hyperbolas even when the problem is restricted to the right branch
  - 原因: The sec-tan form can produce negative $x$ values that are outside the problem scope, leading to invalid solutions
  - 正确做法: Use the hyperbolic parametric form $x = a \cosh t$, $y = b \sinh t$ for problems restricted to the right branch of the hyperbola
- **错误做法:** Forgetting that hyperbolas have asymptotes, while ellipses do not
  - 原因: Asymptotes are often required for hyperbola problems, and omitting them will lose marks
  - 正确做法: Recall that hyperbola asymptotes are $y = \pm \frac{b}{a}x$, given in the formula booklet

## 速查表

| Conic | Standard Cartesian | Parametric Equations | Eccentricity | Foci | Directrices | Tangency Condition $y=mx+c$ | Asymptotes |
| --- | --- | --- | --- | --- | --- | --- | --- |
| Ellipse ($a>b$) | $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ | $x=a \cos t$, $y=b \sin t$ | $e < 1$, $b^2 = a^2(1-e^2)$ | $(\pm ae, 0)$ | $x = \pm \frac{a}{e}$ | $c^2 = a^2 m^2 + b^2$ | None |
| Hyperbola | $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ | $x=a \sec t, y=b \tan t$ / $x=a \cosh t, y=b \sinh t$ | $e > 1$, $b^2 = a^2(e^2 - 1)$ | $(\pm ae, 0)$ | $x = \pm \frac{a}{e}$ | $c^2 = a^2 m^2 - b^2$ | $y = \pm \frac{b}{a}x$ |

## 下一步

Now that you have mastered Further Coordinate Systems for FP3, you can apply these skills to other areas of the Edexcel IAL Further Maths specification. Conic sections often appear alongside differentiation and integration problems, so solidifying your understanding of these coordinate systems will help you tackle more complex calculus questions in FP3. The next topic in FP3 is Further Matrix Algebra, which builds on FP1 matrix content to cover eigenvalues, eigenvectors, and diagonalization, with applications to transformations of conic sections. You should also practice past paper questions on this topic to familiarize yourself with the exam style, as questions often combine multiple parts (eccentricity, tangents, loci) in a single 8-12 mark question. Make sure you can recall the tangency conditions without reference to notes, as these are not provided in the formula book.

- [FP1 Coordinate Systems Recap](https://www.owlsprep.com/zh/study/edexcel-ial-further-math-fp1-coordinate-systems/)

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来自 [OwlsPrep](https://www.owlsprep.com) —— A-Level / IB / AP / IGCSE 免费学习指南，依据官方考纲编写。原页面：https://www.owlsprep.com/zh/study/edexcel-ial-further-math-fp3-further-coordinate-systems/