Parametric Equations for IGCSE Maths: Key Concepts Explained | IGCSE 数学：参数方程 考点精讲

📚 Parametric Equations for IGCSE Maths: Key Concepts Explained | IGCSE 数学：参数方程 考点精讲

Parametric equations appear in IGCSE Additional Mathematics (0606) and offer a powerful way to describe curves using a third variable, usually t or θ, called the parameter. Instead of writing y directly in terms of x, both x and y are expressed as functions of t. This approach helps model motion, circles, ellipses and many other curves that are difficult to describe with a single Cartesian equation. Mastering parametric equations means you can convert between parametric and Cartesian forms, sketch curves, find gradients using differentiation, and solve problems involving tangents and normals. This article covers all the key skills you need for the IGCSE exam, with clear step-by-step explanations.

参数方程出现在 IGCSE 附加数学（0606）中，它使用第三个变量（通常是 t 或 θ，称为参数）来描述曲线。参数方程不是直接写出 y 关于 x 的表达式，而是把 x 和 y 都表示成 t 的函数。这种方法有助于对运动轨迹、圆、椭圆以及许多难以用单个笛卡尔方程描述的曲线进行建模。掌握参数方程意味着你要能够在参数形式和笛卡尔形式之间转换、绘制曲线、用微分求梯度，以及解决切线和法线相关的问题。本文将囊括 IGCSE 考试中所需的所有关键技能，并附有清晰的分步讲解。

1. What Are Parametric Equations? | 什么是参数方程？

In a standard Cartesian equation, we write y = f(x). In parametric form, we write both x = f(t) and y = g(t) for some parameter t. As t varies, the point (x, y) traces out a curve. Parametric equations are extremely useful for describing paths where x and y change independently over time or angle, such as the motion of a projectile or points on a rotating wheel.

在标准的笛卡尔方程中，我们写成 y = f(x)。而在参数形式中，我们对某个参数 t 写成 x = f(t) 和 y = g(t)。随着 t 变化，点 (x, y) 就会描绘出一条曲线。参数方程在描述 x 和 y 随时间或角度独立变化的路径（例如抛体运动或转轮上的点）时极其有用。

For example, the equations x = 2t, y = t² define a parabola. When t = 0, the point is (0, 0); when t = 1, it is (2, 1); when t = 2, it is (4, 4). By plotting several points and joining them smoothly, the shape becomes clear.

例如，方程 x = 2t, y = t² 定义了一条抛物线。当 t = 0 时，点为 (0, 0)；当 t = 1 时，点为 (2, 1)；当 t = 2 时，点为 (4, 4)。描出若干个点并将它们平滑连接后，曲线的形状就清晰可见了。

---

2. Converting Parametric Equations to Cartesian Form | 将参数方程转化为笛卡尔形式

To find the Cartesian equation, eliminate the parameter t between the two equations. This often involves solving one equation for t and substituting into the other, or using trigonometric identities when sin θ and cos θ are involved.

要找到笛卡尔方程，需要在两个方程之间消去参数 t。这通常包括从其中一个方程解出 t 并代入另一个方程，或者当涉及 sin θ 和 cos θ 时利用三角恒等式。

Case 1: x = 3t + 1, y = 2t − 5. Solve for t from x: t = (x − 1)/3. Substitute into y: y = 2((x − 1)/3) − 5 = (2x − 2)/3 − 5 = (2x − 2 − 15)/3 = (2x − 17)/3. So the Cartesian equation is y = (2x − 17)/3, a straight line.

情况一：x = 3t + 1, y = 2t − 5。从 x 中解出 t：t = (x − 1)/3。代入 y：y = 2((x − 1)/3) − 5 = (2x − 2)/3 − 5 = (2x − 2 − 15)/3 = (2x − 17)/3。因此笛卡尔方程为 y = (2x − 17)/3，这是一条直线。

Case 2: x = 2 cos θ, y = 3 sin θ. Use the identity cos²θ + sin²θ = 1. Write (x/2)² + (y/3)² = cos²θ + sin²θ = 1, giving x²/4 + y²/9 = 1, which is an ellipse.

情况二：x = 2 cos θ, y = 3 sin θ。利用恒等式 cos²θ + sin²θ = 1。写出 (x/2)² + (y/3)² = cos²θ + sin²θ = 1，即 x²/4 + y²/9 = 1，这是一个椭圆。

Sometimes a substitution or algebraic manipulation is needed, like squaring both sides when t appears as √t, but be careful with domain restrictions.

有时需要进行代换或代数处理，例如当 t 以 √t 形式出现时需要两边平方，但要注意定义域的限制。

---

3. Sketching Parametric Curves | 绘制参数曲线

To sketch a curve defined by parametric equations, first create a table of values for t, x, and y. Choose values of t that cover the domain and plot the resulting (x, y) coordinates. Pay attention to the direction of the curve as t increases – this is often indicated by arrows on the sketch. IGCSE exam questions may ask you to plot the curve or to identify key features like intercepts and turning points.

要绘制由参数方程定义的曲线，首先制作一个关于 t、x 和 y 的取值表。选取能覆盖定义域的 t 值，并描出对应的 (x, y) 坐标。留意随着 t 增大曲线的走向——通常用箭头在草图上标出。IGCSE 考题可能会要求你画出曲线，或标出截距、转折点等关键特征。

Consider x = t² − 1, y = t³ − t for t from −2 to 2. Calculate points: t=−2 ⇒ (3, −6); t=−1 ⇒ (0, 0); t=0 ⇒ (−1, 0); t=1 ⇒ (0, 0); t=2 ⇒ (3, 6). Plotting these reveals a loop that passes through the origin twice. The direction: as t goes from −2 to 2, the curve moves from (3, −6) upward through (0,0) to (3,6), with a crossing at (−1,0) for t=0.

考虑 x = t² − 1, y = t³ − t，t 从 −2 到 2。计算各点：t=−2 ⇒ (3, −6)；t=−1 ⇒ (0, 0)；t=0 ⇒ (−1, 0)；t=1 ⇒ (0, 0)；t=2 ⇒ (3, 6)。描点后会看到一条经过原点两次的环形曲线。方向：随着 t 从 −2 增大到 2，曲线从 (3, −6) 向上经过 (0,0) 到达 (3,6)，并在 t=0 时经过 (−1,0)。

---

4. Finding the Gradient: dy/dx in Parametric Form | 求梯度：参数形式的 dy/dx

One of the most important skills for IGCSE is differentiating parametric equations to find gradients. The gradient of a curve is given by dy/dx. Using the chain rule, dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. So first find dx/dt and dy/dt separately, then divide.

IGCSE 考试中最重要的技能之一是对参数方程求导以确定梯度。曲线的梯度由 dy/dx 给出。运用链式法则，dy/dx = (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0。因此先分别求出 dx/dt 和 dy/dt，然后相除。

Example: x = t² + 2t, y = t³ − 3t. Then dx/dt = 2t + 2, dy/dt = 3t² − 3. So dy/dx = (3t² − 3)/(2t + 2) = 3(t² − 1)/[2(t + 1)] = 3(t − 1)(t + 1)/[2(t + 1)] = 3(t − 1)/2, for t ≠ −1. At t = 2, gradient = 3(2 − 1)/2 = 3/2.

例如：x = t² + 2t, y = t³ − 3t。那么 dx/dt = 2t + 2, dy/dt = 3t² − 3。因此 dy/dx = (3t² − 3)/(2t + 2) = 3(t² − 1)/[2(t + 1)] = 3(t − 1)(t + 1)/[2(t + 1)] = 3(t − 1)/2，其中 t ≠ −1。当 t = 2 时，梯度 = 3(2 − 1)/2 = 3/2。

Always simplify the gradient expression if possible. It will often be a function of t. To find the gradient at a particular point, first find the t-value at that point by solving the parametric equations if necessary.

尽可能将梯度表达式化简。它通常会是 t 的函数。若要在某一点求梯度，如有必要，先通过解参数方程找到该点对应的 t 值。

---

5. Finding the Equation of Tangents and Normals | 求切线和法线方程

Once you have dy/dx, you can find the equation of the tangent or normal at a specific point. The tangent has gradient m = dy/dx evaluated at that t. The normal has gradient −1/m, provided m ≠ 0. Then use the point-slope form y − y₁ = m(x − x₁) with the point (x(t), y(t)).

一旦有了 dy/dx，就可以求出特定点处的切线或法线方程。切线在该 t 值处的梯度为 m = dy/dx。如果 m ≠ 0，法线的梯度则为 −1/m。然后用点斜式 y − y₁ = m(x − x₁)，结合点 (x(t), y(t)) 即可。

For example, with x = 2t, y = t² + 4, find the tangent at t = 1. Here (x₁, y₁) = (2, 5). dx/dt = 2, dy/dt = 2t ⇒ dy/dx = (2t)/2 = t. At t = 1, m = 1. Tangent: y − 5 = 1(x − 2) ⇒ y = x + 3. Normal: gradient = −1, so y − 5 = −1(x − 2) ⇒ y = −x + 7.

例如，对于 x = 2t, y = t² + 4，求 t = 1 处的切线。此处 (x₁, y₁) = (2, 5)。dx/dt = 2, dy/dt = 2t ⇒ dy/dx = (2t)/2 = t。在 t = 1 处，m = 1。切线：y − 5 = 1(x − 2) ⇒ y = x + 3。法线：梯度 = −1，所以 y − 5 = −1(x − 2) ⇒ y = −x + 7。

---

6. Second Derivative and Concavity | 二阶导数与凹凸性

To determine concavity or locate points of inflection, you need the second derivative d²y/dx². Using parametric differentiation: d²y/dx² = d(dy/dx)/dx = [d(dy/dx)/dt] / (dx/dt). This means differentiate the expression for dy/dx with respect to t, then divide by dx/dt.

要判断凹凸性或确定拐点，需要用到二阶导数 d²y/dx²。使用参数微分法：d²y/dx² = d(dy/dx)/dx = [d(dy/dx)/dt] / (dx/dt)。这意味着先对 dy/dx 的表达式关于 t 求导，然后再除以 dx/dt。

Example: x = t³, y = t². Then dx/dt = 3t², dy/dt = 2t, so dy/dx = 2t/(3t²) = 2/(3t). Then d(dy/dx)/dt = d(2/(3t))/dt = −2/(3t²). Finally, d²y/dx² = [−2/(3t²)] / (3t²) = −2/(9t⁴). This second derivative can help determine if the curve is concave up or down.

例如：x = t³, y = t²。则 dx/dt = 3t², dy/dt = 2t，因此 dy/dx = 2t/(3t²) = 2/(3t)。接着 d(dy/dx)/dt = d(2/(3t))/dt = −2/(3t²)。最后，d²y/dx² = [−2/(3t²)] / (3t²) = −2/(9t⁴)。这个二阶导数有助于判断曲线的凹凸方向。

---

7. Parametric Equations of Circles and Ellipses | 圆和椭圆的参数方程

Standard parametric forms appear frequently in IGCSE:
A circle with radius r centred at the origin: x = r cos θ, y = r sin θ.
A circle centred at (a, b): x = a + r cos θ, y = b + r sin θ.
An ellipse centred at the origin: x = a cos θ, y = b sin θ.
These are vital for modelling periodic motion and geometric problems.

标准参数形式在 IGCSE 中频繁出现：
以原点为圆心、半径为 r 的圆：x = r cos θ, y = r sin θ。
圆心在 (a, b) 的圆：x = a + r cos θ, y = b + r sin θ。
中心在原点的椭圆：x = a cos θ, y = b sin θ。
这些对于模拟周期性运动和几何问题至关重要。

To check: x² + y² = r² cos²θ + r² sin²θ = r². For ellipse: x²/a² + y²/b² = 1. You can also describe half-circles or arcs by restricting the domain of θ. For instance, x = 5 cos θ, y = 5 sin θ for 0 ≤ θ ≤ π gives the upper semicircle.

检验一下：x² + y² = r² cos²θ + r² sin²θ = r²。对于椭圆：x²/a² + y²/b² = 1。你也可以通过限制 θ 的范围来描述半圆或圆弧。例如，x = 5 cos θ, y = 5 sin θ，0 ≤ θ ≤ π，就给出了上半圆。

---

8. Parametric Equations Involving Trigonometric Functions | 涉及三角函数的参数方程

When a parameter is given as θ, both x and y are expressed in terms of sin θ and cos θ. To eliminate the parameter, use Pythagorean identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, etc. Sometimes you need to rewrite x and y to isolate sin θ or cos θ before squaring and adding.

当参数给定为 θ 时，x 和 y 都用 sin θ 和 cos θ 来表示。要消去参数，可利用勾股恒等式：sin²θ + cos²θ = 1，1 + tan²θ = sec²θ 等。有时需要先改写 x 和 y，分离出 sin θ 或 cos θ，然后再平方相加。

Example: x = 3 sec θ, y = 2 tan θ. Write x/3 = sec θ ⇒ (x/3)² = sec²θ; y/2 = tan θ ⇒ (y/2)² = tan²θ. Then sec²θ − tan²θ = 1 gives x²/9 − y²/4 = 1, a hyperbola.

例如：x = 3 sec θ, y = 2 tan θ。写成 x/3 = sec θ ⇒ (x/3)² = sec²θ；y/2 = tan θ ⇒ (y/2)² = tan²θ。然后由 sec²θ − tan²θ = 1 可得 x²/9 − y²/4 = 1，这是一条双曲线。

When finding dy/dx for trigonometric parametrics, compute dx/dθ and dy/dθ, then dy/dx = (dy/dθ)/(dx/dθ). Remember derivatives: d(sin θ)/dθ = cos θ, d(cos θ)/dθ = −sin θ, d(tan θ)/dθ = sec²θ.

求三角函数参数方程的 dy/dx 时，先计算 dx/dθ 和 dy/dθ，然后 dy/dx = (dy/dθ)/(dx/dθ)。记住导数：d(sin θ)/dθ = cos θ，d(cos θ)/dθ = −sin θ，d(tan θ)/dθ = sec²θ。

---

9. Solving Problems Involving Intersections | 解交点问题

To find where a parametric curve meets a line or another curve, substitute the parametric expressions for x and y into the Cartesian equation of the other curve. This yields an equation in t (or θ). Solve for t, then find the corresponding (x, y) coordinates.

要找出参数曲线与一条直线或另一条曲线的交点，可以将 x 和 y 的参数表达式代入另一条曲线的笛卡尔方程中。这样会得到一个关于 t（或 θ）的方程。解出 t，然后求出对应的 (x, y) 坐标。

Example: Find where x = t², y = 2t intersects the line y = x − 3. Substitute: 2t = t² − 3 ⇒ t² − 2t − 3 = 0 ⇒ (t − 3)(t + 1) = 0. t = 3 gives (9, 6); t = −1 gives (1, −2). Both points lie on the line.

例如：求 x = t², y = 2t 与直线 y = x − 3 的交点。代入：2t = t² − 3 ⇒ t² − 2t − 3 = 0 ⇒ (t − 3)(t + 1) = 0。t = 3 对应 (9, 6)；t = −1 对应 (1, −2)。这两点都在该直线上。

Sometimes the resulting equation involves trigonometric functions. Use identities and standard solutions within the given interval to find all possible θ values.

有时得到的方程会包含三角函数。利用恒等式以及给定区间内的标准解法，找出所有可能的 θ 值。

---

10. Domain and Range from Parametric Equations | 参数方程的定义域与值域

The domain of the Cartesian relation is the set of x-values produced as t varies over its allowed interval. The range is the set of corresponding y-values. You may be asked to state the domain and range of the Cartesian equation derived from parametric equations, especially when the parameter is restricted.

笛卡尔关系的定义域，是当 t 在其允许区间内变化时所产生的一组 x 值。值域则是相应 y 值的集合。考题可能会要求你说出从参数方程推导出的笛卡尔方程的定义域和值域，尤其是在参数受到限制时。

For x = 2 cos θ, y = sin θ with 0 ≤ θ ≤ π, x ranges from 2 cos 0 = 2 to 2 cos π = −2, so domain is −2 ≤ x ≤ 2. However, because sin θ is non-negative on [0,π], y ranges from 0 to 1, not the full −1 ≤ y ≤ 1. So the range is 0 ≤ y ≤ 1. Always check endpoints and turning points inside the interval.

对于 x = 2 cos θ, y = sin θ，且 0 ≤ θ ≤ π，x 的范围是从 2 cos 0 = 2 到 2 cos π = −2，因此定义域为 −2 ≤ x ≤ 2。然而，由于 sin θ 在 [0,π] 上非负，y 的范围是 0 到 1，而不是完整的 −1 ≤ y ≤ 1。因此值域为 0 ≤ y ≤ 1。务必检查区间内的端点和转折点。

---

11. Parametric Differentiation: Common Pitfalls | 参数微分：常见陷阱

– Forgetting to use the chain rule: dy/dx is not just dy/dt. Always divide by dx/dt.
– Mixing up the formula for the second derivative: d²y/dx² is not d²y/dt² divided by d²x/dt². You must differentiate dy/dx with respect to t first, then divide by dx/dt.
– Not simplifying dy/dx properly before differentiating again.
– Losing domain restrictions when squaring equations during elimination.

– 忘记使用链式法则：dy/dx 不仅仅是 dy/dt。始终要除以 dx/dt。
– 把二阶导数的公式搞混：d²y/dx² 不是 d²y/dt² 除以 d²x/dt²。你必须先对 dy/dx 关于 t 求导，然后再除以 dx/dt。
– 在再次求导之前没有充分化简 dy/dx。
– 在消元过程中进行平方运算时丢失了定义域限制。

Be methodical: write down dx/dt and dy/dt clearly, then form dy/dx and simplify. For the second derivative, label it as d²y/dx² and work step by step.

要有条不紊：清晰地写出 dx/dt 和 dy/dt，然后构成 dy/dx 并化简。对于二阶导数，将其标记为 d²y/dx² 并逐步运算。

---

12. Exam-Style Question and Summary | 考试风格题目与总结

Question: A curve is defined parametrically by x = 3 sin θ, y = 4 cos θ + 1, for 0 ≤ θ ≤ 2π.
(a) Find the Cartesian equation of the curve.
(b) Determine dy/dx in terms of θ.
(c) Find the equations of the tangent and normal at θ = π/6.
(d) State the domain and range of the Cartesian equation.

题目：一条曲线由参数方程定义为 x = 3 sin θ, y = 4 cos θ + 1，其中 0 ≤ θ ≤ 2π。
(a) 求该曲线的笛卡尔方程。
(b) 用 θ 表示 dy/dx。
(c) 求 θ = π/6 处的切线和法线方程。
(d) 说出笛卡尔方程的定义域和值域。

Solutions:
(a) sin θ = x/3, cos θ = (y − 1)/4. Squaring and adding: x²/9 + (y − 1)²/16 = 1, an ellipse.
(b) dx/dθ = 3 cos θ, dy/dθ = −4 sin θ ⇒ dy/dx = (−4 sin θ)/(3 cos θ) = (−4/3) tan θ.
(c) At θ = π/6: sin(π/6)=1/2, cos(π/6)=√3/2. x = 3(1/2)=3/2, y=4(√3/2)+1=2√3+1. Gradient m = (−4/3)(1/√3) = −4/(3√3) = −4√3/9. Tangent: y − (2√3+1) = (−4√3/9)(x − 3/2). Normal: gradient = 9/(4√3) = 3√3/4, then use point-slope form.
(d) Since sin θ and cos θ both range [−1,1], x ∈ [−3,3], y ∈ [−3,5] because 4 cos θ ∈ [−4,4] then +1 gives [−3,5].

解答：
(a) sin θ = x/3，cos θ = (y − 1)/4。平方相加得：x²/9 + (y − 1)²/16 = 1，这是一个椭圆。
(b) dx/dθ = 3 cos θ，dy/dθ = −4 sin θ ⇒ dy/dx = (−4 sin θ)/(3 cos θ) = (−4/3) tan θ。
(c) 在 θ = π/6 处：sin(π/6)=1/2，cos(π/6)=√3/2。x = 3(1/2)=3/2，y=4(√3/2)+1=2√3+1。梯度 m = (−4/3)(1/√3) = −4/(3√3) = −4√3/9。切线：y − (2√3+1) = (−4√3/9)(x − 3/2)。法线：梯度 = 9/(4√3) = 3√3/4，然后使用点斜式。
(d) 由于 sin θ 和 cos θ 的范围都是 [−1,1]，x ∈ [−3,3]，y ∈ [−3,5]，因为 4 cos θ ∈ [−4,4] 加 1 后得到 [−3,5]。

Mastering parametric equations gives you a robust set of tools for tackling IGCSE questions involving curves, motion and geometry. Practice converting forms, differentiating, and interpreting geometric properties to score full marks.

掌握参数方程为你提供了一套有力的工具，以应对 IGCSE 考试中涉及曲线、运动与几何的问题。通过练习形式转换、微分以及解读几何性质，你就能拿下满分。

Published by TutorHao | IGCSE Additional Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）