📚 Example 6.12.2: Tangent to a Parametric Curve | 示例 6.12.2：参数方程曲线的切线
This worked example focuses on finding the tangent to a curve defined by parametric equations. It is a common question in the AQA A-Level Mathematics pure specification and tests your ability to handle parametric differentiation, evaluate derivatives, and construct a straight-line equation.
本例题重点讲解如何求由参数方程定义的曲线的切线。这是 AQA A-Level 数学纯数部分常见题型，考查参数求导、导数计算以及直线方程的建立。
1. The Worked Example | 例题展示
where t = 2. Give your answer in the form ax + by + c = 0, where a, b and c are integers.
曲线 C 由参数方程 x = t² + 1 和 y = t³ − 3t 给出，其中 t 为实数参数。求曲线 C 在 t = 2 处的切线方程，结果写成 ax + by + c = 0 的形式，a、b、c 为整数。
2. Understanding Parametric Curves | 理解参数曲线
f(x), so you must use parametric differentiation rather than trying to eliminate t first.
参数曲线通过第三个变量 t 分别描述 x 和 y。随着 t 变化，点 (x, y) 描绘出一条曲线。这种表示可以描述不是 y = f(x) 形式的曲线，因此必须使用参数求导，而不是先消去 t。
The tangent at a point gives the best linear approximation to the curve at that point. Finding its equation requires two pieces of information: the coordinates of the point and the gradient of the curve at that point.
曲线上某点的切线给出了该点处曲线的最佳线性近似。求切线方程需要两个信息：该点的坐标和曲线在该点的斜率。
3. Step 1: Find the Point | 第1步：确定切点坐标
. For y, calculate y = 2³ − 3 × 2 = 8 − 6 = 2. Therefore the point of tangency is P(5, 2).
将 t = 2 代入参数方程。对于 x，计算 x = 2² + 1 = 5；对于 y，计算 y = 2³ − 3 × 2 = 8 − 6 = 2。因此切点坐标为 P(5, 2)。
4. Step 2: Differentiate Parametrically | 第2步：参数求导
tiate x with respect to t: dx/dt = 2t. Differentiate y with respect to t: dy/dt = 3t² − 3.
要求 dy/dx，使用链式法则形式 dy/dx = (dy/dt) / (dx/dt)。对 x 关于 t 求导：dx/dt = 2t；对 y 关于 t 求导：dy/dt = 3t² − 3。
dy/dx = (dy/dt) / (dx/dt)
/dx)(dx/dt) provided dx/dt is not zero. Rearranging gives the parametric form of the derivative, which is essential when a curve crosses itself or is not a function.
该公式成立是因为 dy/dt = (dy/dx)(dx/dt)，只要 dx/dt 不为零。整理即可得到导数的参数形式。当曲线自身相交或不是函数时，这一方法必不可少。
5. Step 3: Evaluate the Gradient | 第3步：计算切线斜率
luate dx/dt and dy/dt. Compute dx/dt = 2 × 2 = 4. Compute dy/dt = 3 × 2² − 3 = 12 − 3 = 9.
在 t = 2 处计算 dx/dt 和 dy/dt。计算 dx/dt = 2 × 2 = 4；计算 dy/dt = 3 × 2² − 3 = 12 − 3 = 9。
to the parametric derivative: dy/dx = 9/4. This is the gradient of the tangent at P(5, 2).
现在代入参数导数：dy/dx = 9/4。这就是 P(5, 2) 处切线的斜率。
dy/dx = 9/4
6. Step 4: Build the Tangent Equation | 第4步：建立切线方程
-gradient form of a straight line: y − y₁ = m(x − x₁). Here m = 9/4 and (x₁, y₁) = (5, 2).
使用直线的点斜式：y − y₁ = m(x − x₁)。这里 m = 9/4，(x₁, y₁) = (5, 2)。
y − 2 = (9/4)(x − 5)
d form ax + by + c = 0. Multiply every term by 4 to clear the fraction before rearranging.
展开并化简，将方程写成题目要求的 ax + by + c = 0 形式。先将每一项乘以 4 以消去分母，再整理。
7. Simplifying the Answer | 化简答案
). Expand: 4y − 8 = 9x − 45. Rearrange to collect all terms on one side: 9x − 4y − 37 = 0.
两边乘以 4：4(y − 2) = 9(x − 5)。展开：4y − 8 = 9x − 45。移项，将所有项移到同一边：9x − 4y − 37 = 0。
the tangent equation is 9x − 4y − 37 = 0 with integer coefficients a = 9, b = −4, c = −37.
因此切线方程为 9x − 4y − 37 = 0，其中整数系数 a = 9，b = −4，c = −37。
8. Verifying the Result | 验证结果
to 9x − 4y − 37. Compute 9 × 5 − 4 × 2 − 37 = 45 − 8 − 37 = 0. The point lies on the line.
验证点 P(5, 2) 是否满足方程：将 x = 5 和 y = 2 代入 9x − 4y − 37。计算 9 × 5 − 4 × 2 − 37 = 45 − 8 − 37 = 0。该点在直线上。
be rearranged as y = (9/4)x − 37/4, so its gradient is 9/4, which matches dy/dx at t = 2.
还需检查斜率是否一致。直线 9x − 4y − 37 = 0 可改写为 y = (9/4)x − 37/4，斜率为 9/4，与 t = 2 时的 dy/dx 相符。
9. Common Mistakes | 常见错误
ing the parameter, which may fail or create confusion because the Cartesian form may involve roots or restricted domains. Always differentiate parametrically unless the parameter is very easy to eliminate.
一个常见错误是消去参数后直接将 t = 2 代入 y = f(x)，这可能会失败或造成混淆，因为笛卡尔形式可能含有根号或受限的定义域。除非参数非常容易消去，否则应始终使用参数求导。
− 3)/(2t) and then use this general expression as the gradient without substituting t = 2.
另一个错误是忘记在给定 t 值处计算导数。学生有时写出 dy/dx = (3t² − 3)/(2t) 后，不代入 t = 2 就将这个一般表达式当作斜率。
change ax + by + c = 0 into an incorrect line, even if the gradient and point are correct.
在整理最终方程时还要注意符号。符号错误会把 ax + by + c = 0 变成错误的直线，即使斜率和点都正确。
10. Exam Tips and Extension | 考试技巧与拓展
In an exam, always write down dx/dt and dy/dt before dividing. Showing your working earns method marks even if a numerical error occurs later. Also state the parametric derivative formula clearly.
考试时，一定要先写出 dx/dt 和 dy/dt 再做除法。即使后续出现数值错误，展示解题过程也能获得方法分。还要清楚地写出参数导数公式。
This technique extends to finding tangents and normals for parametric curves. For a normal, use the negative reciprocal of the gradient: mnormal = −1 / (dy/dx).
这一方法可以拓展到求参数曲线的切线和法线。对于法线，使用斜率的负倒数：mnormal = −1 / (dy/dx)。
If a question asks for the point where a tangent is parallel to a given line, set dy/dx equal to that line's gradient and solve for t. Then substitute back to find the coordinates. This links parametric differentiation with coordinate geometry.
如果题目问切线平行于给定直线的位置，就令 dy/dx 等于该直线的斜率并解出 t，然后代回求坐标。这就把参数求导与坐标几何联系起来了。
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