Handle horizontal vertical and indeterminate tangent cases
Use coordinate rates or implicit derivative coefficients to recognise regular vertical and horizontal tangents, and investigate a zero-over-zero ratio separately.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.
Before you begin
- Find parametric and implicit first derivatives.
- Represent horizontal and vertical lines and recognise an undefined ratio.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does a derivative ratio 0/0 tell you by itself?
What you will learn
- Find tangent and normal equations when a regular curve has a horizontal or vertical tangent.
- Distinguish a non-zero-over-zero gradient from an indeterminate zero-over-zero parametrisation and inspect the original relation.
A regular curve direction can be vertical without a finite gradient
For a differentiable parametric curve, if dx/dt ≠ 0 and dy/dt = 0, the tangent is horizontal and its normal vertical. If dx/dt = 0 and dy/dt ≠ 0, the local direction is vertical: the tangent is x = x₀ and the normal y = y₀. Do not call the gradient zero just because its denominator is zero, or force a vertical tangent into y = mx + c. A non-zero coordinate-rate vector supplies a definite local direction even when dy/dx has no finite value.
On x = cos t, y = sin t, at t = 0 the point is (1, 0) and the rates are 0 and 1. Thus the tangent is x = 1 and the normal y = 0. At t = π/2 the point is (0, 1) and the rates are −1 and 0, giving tangent y = 1 and normal x = 0. The same geometry follows from x² + y² = 1: its formula y′ = −x/y is finite at the top point but needs a separate vertical-tangent check at y = 0.
Zero over zero does not decide a curve's tangent
Implicit differentiation can give a relation A(x, y) + B(x, y)y′ = 0. If B ≠ 0, divide to find y′. If B = 0 and A ≠ 0 at a regular curve point, the finite y-as-a-function-of-x slope fails and the tangent is vertical. If both coefficients vanish, the first derivative relation alone is inconclusive: inspect the original relation or its local branches. Likewise, when both parametric coordinate rates are zero, their 0/0 ratio does not automatically mean a zero slope, no tangent or a unique vertical tangent.
For x = t³ and y = 2t³, both rates vanish at t = 0. Eliminating the parameter gives y = 2x for every real x, so the curve is a straight line with tangent y = 2x and normal y = −x/2 at the origin. The chosen parametrisation stops changing to first order at zero, but the underlying curve still has gradient 2. Recovering the original geometric relation supplies evidence that the indeterminate ratio alone could not provide.
Pause and explain
At a regular parametric point, dx/dt = 0 and dy/dt = 2. What is the tangent direction?
Worked example
For x = cos t, y = sin t, find tangent and normal equations at t = 0 and t = π/2 without dividing by a zero rate.
Show the worked solution
- At t = 0 the point is (1, 0), with dx/dt = 0 and dy/dt = 1; its tangent is vertical x = 1.
- The normal there is horizontal y = 0.
- At t = π/2 the point is (0, 1), with rates −1 and 0; its tangent is horizontal y = 1 and its normal vertical x = 0.
Answer At t = 0: tangent x = 1, normal y = 0. At t = π/2: tangent y = 1, normal x = 0.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For x² + y² = 25, find tangent and normal equations at (5, 0) and explain why y′ = −x/y cannot give a finite gradient there.
Give me a hint
Verify the point and inspect the original circle geometry or its non-zero implicit x coefficient.
Compare my reasoning
- The point satisfies the circle equation; its derivative relation is 2x + 2y y′ = 0.
- At (5, 0), the y′ coefficient is zero while the x contribution is 10, so the regular tangent is vertical x = 5.
- The perpendicular normal is horizontal y = 0; the invalid ratio −5/0 should not be replaced by zero.
Tangent x = 5; normal y = 0; no finite dy/dx at that point.
Look for these in your work
- Distinguish an original valid point from an undefined finite slope formula.
- Use vertical and horizontal line equations instead of dividing by zero.
For x = t² and y = t, find tangent and normal equations at t = 0 and check by eliminating t.
Give me a hint
The vertical coordinate rate stays non-zero even where the horizontal rate vanishes.
Compare my reasoning
- At zero the point is (0, 0), with dx/dt = 2t = 0 and dy/dt = 1.
- The tangent is vertical x = 0 and the normal horizontal y = 0.
- Eliminating t gives x = y², whose sideways parabola has the same vertical tangent at its origin.
Tangent x = 0; normal y = 0; Cartesian check x = y².
Look for these in your work
- Use the non-zero local direction vector despite a zero horizontal rate.
- Confirm the tangent using the original geometric relation.
For x = t³ and y = 2t³, find the tangent and normal at t = 0. Explain why substituting into (dy/dt)/(dx/dt) is insufficient.
Give me a hint
Both coordinate rates vanish; eliminate the parameter to identify the underlying curve.
Compare my reasoning
- The rates are 3t² and 6t², so direct substitution gives 0/0 rather than a defined gradient.
- The original coordinates satisfy y = 2x for every real x, proving that the curve is a line with gradient 2 through the origin.
- Its tangent is y = 2x and its perpendicular normal is y = −x/2; the explicit relation resolves the indeterminate parametrisation.
Tangent y = 2x; normal y = −x/2; the 0/0 parameter ratio alone is inconclusive.
Look for these in your work
- Recognise the indeterminate ratio without assigning it a value.
- Use a verified equivalent curve relation to find the tangent and normal.
Common mistakes
- A zero denominator with a non-zero numerator is not a zero gradient.
- Use x = constant for a vertical line instead of assigning it a finite gradient.
- A 0/0 ratio is inconclusive; use the original relation or nearby curve behaviour for a separate argument.
What can you explain now?
For x = t³, y = 2t³ at t = 0, does the 0/0 derivative ratio prove that there is no tangent?
Compare with the explanation
No; eliminating t gives y = 2x, so the tangent at the origin is y = 2x.
The parametrisation's zero rates do not determine the underlying curve's tangent; its explicit line relation does.
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Make it stick
Compare horizontal, vertical and indeterminate derivative cases. Give a valid curve point where the finite slope formula fails and explain which separate geometric evidence resolves the tangent.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.