A tangent gives two facts at the point of contact: the curve and the line have the same y-value, and the same gradient. Each fact becomes one equation, which lets you solve for unknown constants in the curve.
This lesson is part of recovering the method in a mixed calculus problem. It builds on finding tangents and normals.
Worked example 1: one unknown from a gradient
The curve y = ax² + 3x − 1 has gradient 11 at x = 2. Find a.
dy/dx = 2ax + 3. At x = 2 this is 4a + 3, and it equals 11.
So 4a + 3 = 11, giving 4a = 8 and a = 2. Check: dy/dx = 4x + 3 = 11 at x = 2.
Worked example 2: two unknowns from a tangent line
The line y = 2x + 3 is tangent to the curve y = x³ + ax + b at the point where x = 1. Find a and b.
Same gradient. The curve’s gradient is 3x² + a, which equals 3 + a at x = 1. The line’s gradient is 2, so 3 + a = 2 and a = −1.
Same point. The line at x = 1 has y = 2(1) + 3 = 5. The curve at x = 1 has y = 1 + a + b = 1 − 1 + b = b, so b = 5.
Check. The curve is y = x³ − x + 5. At x = 1: y = 1 − 1 + 5 = 5, and dy/dx = 3 − 1 = 2. Both match the line.
Worked example 3: the point is stated
The line y = 4x − 5 is tangent to y = x² + bx + 4 at x = 3.
The line at x = 3 gives y = 12 − 5 = 7. The curve at x = 3 gives 9 + 3b + 4 = 7, so 3b = −6 and b = −2.
The gradient check: the curve’s gradient is 2x + b = 6 + b, which equals 4 when b = −2. The two conditions agree, which confirms the answer.
The mistake that costs marks
The common slip is to use only one of the two conditions, or to use the line’s y-intercept as if it were the curve’s value.
| Step | Wrong | Right |
|---|---|---|
| Point condition | Put the line’s constant (+3) equal to the curve’s constant | Substitute x = 1 into both and set the y-values equal |
| Gradient condition | Set the curve’s y equal to the line’s gradient | Set dy/dx of the curve equal to the line’s gradient |
Always write two separate lines of working, one headed “same gradient” and one headed “same point”.
Check yourself
The line y = 3x − 1 is tangent to the curve y = x² + px + q at the point where x = 2. Find p and q.
Answer
Gradient: dy/dx = 2x + p. At x = 2 it is 4 + p, which equals 3, so p = −1.
Point: the line at x = 2 gives y = 6 − 1 = 5. The curve at x = 2 gives 4 + 2p + q = 4 − 2 + q = 2 + q, so 2 + q = 5 and q = 3.
Check: the curve is y = x² − x + 3. At x = 2: y = 4 − 2 + 3 = 5 and dy/dx = 4 − 1 = 3. Both match.
What to study next
Continue with comparing endpoints and stationary points in a constrained optimisation. For the discriminant route to a similar problem, see a line touching a curve at one point.
To have a teacher check the equations you set up, see online one-to-one Additional Mathematics tuition.