Parallel lines have the same gradient. To find tangents parallel to a given line, differentiate the curve and set its derivative equal to the gradient of that line.
The derivative equation can give more than one x-value. Use each value in the original curve equation to find the point of contact. Then write a tangent equation for each point.
- Read the gradient from the given line.
- Solve the derivative equation for every possible x-value.
- Find each point of contact and substitute it into the tangent equation.
Worked example | Two tangents to one cubic
Find the tangents to y = x³ − 3x that are parallel to y = 9x − 5. The given line has gradient 9. Differentiating the curve gives dy/dx = 3x² − 3.
Set 3x² − 3 = 9. Then x² = 4, so x = 2 or x = −2. The original curve gives y = 2 at x = 2 and y = −2 at x = −2. The points of contact are (2, 2) and (−2, −2).
A line with gradient 9 has the form y = 9x + c. The point (2, 2) gives c = −16. The point (−2, −2) gives c = 16. The two tangents are y = 9x − 16 and y = 9x + 16.
Use the intercept to check each line
The reference line’s intercept −5 plays no part in locating the points. It tells you where the reference line crosses the y-axis. Each tangent gets its own intercept from its point of contact.
A final check is to substitute the x-coordinate into both the curve and the proposed tangent. Then confirm the line’s gradient matches the derivative there.
Try a lower gradient on the same curve
Find all tangents to y = x³ − 3x parallel to y = 3x + 1.
Answer. Solve 3x² − 3 = 3 to get x = ±√2. At either point, x³ = 2x, so y = −x. In y = 3x + c, this gives c = −4x. The tangents are y = 3x − 4√2 and y = 3x + 4√2.
Frequently asked questions
Can there be just one tangent with the required gradient?
Yes. On this cubic, a gradient of −3 gives x = 0 only. A gradient below −3 gives no real solution.
Why substitute into the original curve equation?
The derivative gives the gradient at a point. The original equation gives its y-coordinate.
Official information
SEAB, 2027 SEC G3 Additional Mathematics syllabus (K341), differentiation applied to tangents. Syllabus checked 11 November 2026. The examples are original.