📚 Example 1.5.3: Using the Discriminant for Tangency and Roots | 例 1.5.3：利用判别式求相切与根的条件
x + 5 and the line L: y = kx + 3, and asks for the values of k that make L a tangent to C.
在这个 AQA 风格的工作示例中，我们探讨判别式的一个经典应用：求直线与二次曲线相切时参数 k 的值。题目给出曲线 C: y = x² + 2x + 5 和直线 L: y = kx + 3，要求找出使 L 与 C 相切的 k 值。
This example is representative of the type of question found in AQA A-Level Mathematics papers, combining algebraic manipulation, the discriminant, and geometric interpretation of roots. A clear understanding of this method will help you tackle many intersection and tangency problems.
这个例子代表了 AQA A-Level 数学试卷中常见的一类题，综合了代数运算、判别式以及对根的几何解释。清晰掌握这个方法将帮助你解决许多交点和相切问题。
1. The Core Question in Example 1.5.3 | 例 1.5.3 的核心问题
s a real constant. The objective is to find the values of k such that L is a tangent to C.
我们已知曲线 C: y = x² + 2x + 5 和直线 L: y = kx + 3，其中 k 为实常数。目标是求出使 L 与 C 相切的 k 值。
This is a discriminant question because when we set the two equations equal, the resulting equation is a quadratic in x whose roots correspond to the x-coordinates of the intersection points. A tangent occurs exactly when that quadratic has one repeated root.
这是一道判别式题，因为当我们将两个方程联立时，会得到关于 x 的二次方程，其根对应交点的 x 坐标。相切恰好发生在该二次方程只有一个重根时。
In the AQA specification, this type of problem tests your ability to link algebraic conditions to geometric behaviour, and it appears frequently in both AS and A-Level papers.
在 AQA 考试大纲中，这类问题考查你将代数条件与几何行为联系起来的能力，它在 AS 和 A-Level 试卷中都经常出现。
2. The Discriminant Δ = b² − 4ac | 判别式 Δ = b² − 4ac
s Δ = b² − 4ac. It tells us the nature of the roots without solving the equation directly.
对于实系数二次方程 ax² + bx + c = 0，判别式定义为 Δ = b² − 4ac。它无需直接解方程就能告诉我们根的性质。
The discriminant has three important cases:
判别式有三种重要情况：
| Condition | Nature of roots | Geometry |
|---|---|---|
| Δ > 0 | Two distinct real roots | Two intersections |
| Δ = 0 | One repeated real root | Tangent |
| Δ < 0 | No real roots | No intersection |
In many AQA questions, you are expected to quote this table accurately and then apply it to a specific algebraic equation. Always check that your quadratic is in standard form before reading off a, b, and c.
在许多 AQA 考题中，你需要准确引用这个表格，然后将其应用于具体的代数方程。在读取 a、b、c 之前，务必检查二次方程是否为标准形式。
3. Interpreting the Discriminant in Context | 结合题意解释判别式
When we set the line and the curve equal, the resulting quadratic equation has roots that are precisely the x-coordinates of the intersection points of L and C.
当我们将直线与曲线联立后，所得二次方程的根正是直线 L 与曲线 C 交点的 x 坐标。
If the discriminant is positive, there are two distinct real roots, so the line cuts the curve at two different points. If the discriminant is zero, there is exactly one repeated root, meaning the line just touches the curve at a single point, which is the definition of a tangent.
如果判别式为正，则有两个不同的实根，因此直线与曲线相交于两个不同点。如果判别式为零，则恰好有一个重根，这意味着直线刚好在一点接触曲线，即相切。
If the discriminant is negative, there are no real roots, so the line does not meet the curve at all. This geometric interpretation is essential for understanding what the question is really asking.
如果判别式为负，则没有实根，因此直线与曲线完全没有交点。这种几何解释对于理解题目真正问的是什么至关重要。
4. Setting Up the Intersection Equation | 建立交点方程
To find the intersection points, equate the two expressions for y:
为了求交点，令两个 y 的表达式相等：
kx + 3 = x² + 2x + 5
Rearranging to standard quadratic form gives:
整理成标准二次形式：
x² + (2 − k)x + 2 = 0
k, and c = 2. Be careful with signs when subtracting the terms of the line from the curve.
这里系数为 a = 1，b = 2 − k，c = 2。在将直线项从曲线项中减去时要注意符号。
It is very easy to make a sign error at this stage, so always write out the rearrangement in full. A small mistake in b will change the discriminant and lead to the wrong values of k.
在这一步很容易出现符号错误，因此务必完整写出整理过程。b 的小错误会改变判别式，导致 k 的值错误。
5. Applying the Tangency Condition | 应用相切条件
For the line to be tangent to the curve, the quadratic must have exactly one repeated root. Therefore we require the discriminant to be zero.
要使直线与曲线相切，二次方程必须恰好有一个重根。因此我们要求判别式为零。
Δ = (2 − k)² − 4(1)(2) = 0
Expand the bracket carefully:
仔细展开括号：
(2 − k)² = k² − 4k + 4
Substitute this back into the discriminant equation:
将其代回判别式方程：
k² − 4k + 4 − 8 = 0
Simplify to obtain a quadratic in k:
化简得到关于 k 的二次方程：
k² − 4k − 4 = 0
This is the hidden quadratic that many students expect in this type of question. The parameter k itself satisfies a quadratic equation because the tangency condition is a quadratic constraint.
这是许多学生在同类题中会遇到的隐含二次方程。参数 k 本身满足一个二次方程，因为相切条件是一个二次约束。
6. Solving the Hidden Quadratic in k | 解关于 k 的隐含二次方程
Now solve k² − 4k − 4 = 0 using the quadratic formula with A = 1, B = −4, and C = −4.
现在使用二次公式解 k² − 4k − 4 = 0，其中 A = 1，B = −4，C = −4。
k = [−B ± √B² − 4AC] / 2A
Substitute the values:
代入数值：
k = [4 ± √((−4)² − 4(1)(−4))] / 2
Simplify inside the square root:
化简根号内的表达式：
k = [4 ± √16 + 16] / 2 = [4 ± √32] / 2
Since √32 = 4√2, we can write:
由于 √32 = 4√2，我们可以写成：
k = [4 ± 4√2] / 2 = 2 ± 2√2
Therefore the two gradients for which the line is tangent to the curve are k = 2 + 2√2 and k = 2 − 2√2.
因此，直线与曲线相切时的两个斜率为 k = 2 + 2√2 和 k = 2 − 2√2。
These are exact values, and in an AQA exam you should give answers in this simplified surd form unless the question asks for a decimal approximation.
这些都是精确值，在 AQA 考试中，除非题目要求小数近似，否则你应以这种简化根式形式给出答案。
7. Checking Your Answers by Substitution | 通过代入检查答案
When k = 2 + 2√2, substitute back into the intersection equation x² + (2 − k)x + 2 = 0. Since 2 − k = −2√2, the equation becomes:
当 k = 2 + 2√2 时，代回交点方程 x² + (2 − k)x + 2 = 0。因为 2 − k = −2√2，方程变为：
x² − 2√2 x + 2 = 0
This can be written as (x − √2)² = 0, so it has a repeated root x = √2. Substitute x = √2 into either equation to find the y-coordinate and confirm the point of contact.
这可以写成 (x − √2)² = 0，所以它有重根 x = √2。将 x = √2 代入任一方程即可求出 y 坐标并确认切点。
Similarly, when k = 2 − 2√2, we have 2 − k = 2√2, giving:
类似地，当 k = 2 − 2√2 时，2 − k = 2√2，得到：
x² + 2√2 x + 2 = 0
This is (x + √2)² = 0, giving the repeated root x = −√2. This verification step is a powerful way to check that your discriminant
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