When a line touches a curve at one point, the line and curve meet exactly once. For a quadratic curve, that means putting the two equations equal gives a quadratic with one repeated root, so its discriminant is zero.
This lesson belongs to geometry problems with several possible approaches. It uses the discriminant from quadratic functions and the gradient idea from differentiation.
Route 1: the discriminant
Take the curve y = x² − 4x + 5 and the line y = 2x + k. The line touches the curve at one point, so find k.
- Put the equations equal: x² − 4x + 5 = 2x + k.
- Move everything to one side: x² − 6x + (5 − k) = 0.
- One point of contact means b² − 4ac = 0, so (−6)² − 4(1)(5 − k) = 0.
- Simplify: 36 − 20 + 4k = 0, so 4k = −16 and k = −4.
The brackets around 5 − k matter. Dropping them is a common slip in this type of question.
Route 2: the gradient
The curve’s gradient is dy/dx = 2x − 4. The line’s gradient is 2, and at the point of contact the two gradients are equal.
Set 2x − 4 = 2, so x = 3. Then y = 9 − 12 + 5 = 2 on the curve. The line passes through (3, 2), so 2 = 2(3) + k, which gives k = −4.
Both routes agree. That agreement is your check, and in an exam you can use one route to verify the other if time allows.
When the parameter is the gradient
Now take the line y = mx + 1 and the curve y = x² + 3x + 5. Find the values of m for which the line touches the curve.
Putting them equal gives x² + 3x + 5 = mx + 1, so x² + (3 − m)x + 4 = 0. The discriminant is (3 − m)² − 4(1)(4) = 0, so (3 − m)² = 16.
That means 3 − m = 4 or 3 − m = −4, so m = −1 or m = 7. Both are correct: two different lines through (0, 1) each touch the curve, at x = −2 and x = 2.
The mistake that costs marks
A common slip is to set the discriminant greater than zero when the question says “touches at one point”. That is the condition for two meeting points.
| Wording | Number of common points | Condition |
|---|---|---|
| Line cuts the curve at two points | 2 | b² − 4ac > 0 |
| Line touches the curve | 1 | b² − 4ac = 0 |
| Line does not meet the curve | 0 | b² − 4ac < 0 |
Read the wording first, then pick the condition from the table.
Check yourself
The line y = x + k touches the curve y = x² − 3x + 6. Find k, then find the point of contact.
Answer
Put equal: x² − 3x + 6 = x + k, so x² − 4x + (6 − k) = 0.
Discriminant: 16 − 4(6 − k) = 0, so 16 − 24 + 4k = 0 and k = 2.
With k = 2, the equation is x² − 4x + 4 = 0, so x = 2. Then y = 4 − 6 + 6 = 4. Check on the line: y = 2 + 2 = 4. The point of contact is (2, 4).
What to study next
Move on to checking an algebraic intersection against a stated geometric restriction, then try the mixed geometry practice.
If you would like a teacher to set you fresh tangent questions and watch your method choice, see online one-to-one Additional Mathematics tuition.