01 · A tangent at a negative parameter
For x = t² + 1, y = t³ − 3t, find the tangent at t = −2.
Hint
The point is (5, −2) and the gradient is −9/4.
Worked solution
y + 2 = −(9/4)(x − 5), or 4y = −9x + 37.
Understand · explore · practise
Parametric tangents and normals
Find tangent and normal equations to parametric curves. Practise exact points, horizontal and vertical cases, parallel gradients, second intersections and axis-intercept geometry.
Before you startParametric differentiation; straight-line equations; exact radian values
01 / Start with a point and a direction
A gradient alone does not specify a line.
Tangent: y − y₀ = m(x − x₀)
Normal: y − y₀ = −(1/m)(x − x₀)
The normal formula applies when the tangent gradient m is finite and nonzero. Handle horizontal and vertical tangents separately.
At parameter t = a, calculate (x₀, y₀) from the original equations and m from the ratio of the parameter derivatives. The tangent and normal pass through the same contact point.
Point, tangent and normalExplore
t = 0.5; point = (0.5, 0.25); tangent gradient = 1.
Tangent: y − 0.25 = 1(x − 0.5).
This parabola meets its tangent only at the contact point.
Choose the cubic and a nonzero contact parameter to see the tangent meet the curve again. At t = 0 its horizontal tangent crosses at the point of inflection itself. A tangent need not stay on one side of a curve.
02 / Write a parametric tangent equation
Evaluate the coordinates and gradient at the same parameter.
x = t² + 1, y = t³ − 3t, at t = 2Worked example
Point: (5, 2)
Substitute t = 2 into both coordinates.
dy/dx = (3t² − 3)/(2t), so m = 9/4
Divide the two parameter rates.
y − 2 = (9/4)(x − 5)
Use the contact point, not the parameter, as the x-coordinate.
4y = 9x − 37
An equivalent rearranged line equation.
Watch a tangent meet the cubic at a second point
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Take the negative reciprocal for a normal
Use the tangent gradient before changing the direction.
Normal to x = t² + 1, y = t³ − 3t at t = 2Worked example
Tangent gradient: 9/4
The contact point is still (5, 2).
Normal gradient: −4/9
The product of the two finite gradients is −1.
y − 2 = −(4/9)(x − 5)
Both lines go through the same point.
9y = −4x + 38
Check by substituting (5, 2).
02 · Check a normal
Find the normal to x = t² + 1, y = t³ − 3t at t = −2.
Hint
The tangent gradient is negative, so the normal gradient is positive.
Worked solution
The normal is y + 2 = (4/9)(x − 5), or 9y = 4x − 38.
03 · Exponential coordinates
Find the tangent and normal to x = eᵗ, y = t² at t = 1.
Hint
The point is (e, 1), with tangent gradient 2/e.
Worked solution
Tangent: y − 1 = (2/e)(x − e). Normal: y − 1 = −(e/2)(x − e).
04 / Handle horizontal and vertical lines directly
A vertical line cannot be written with a finite gradient.
Horizontal tangent y = y₀ → vertical normal x = x₀
Vertical tangent x = x₀ → horizontal normal y = y₀
x = t² + 1, y = t³ − 3tWorked example
At t = 1: point (2, −2), dy/dt = 0, dx/dt = 2
Tangent y = −2; normal x = 2.
At t = 0: point (1, 0), dx/dt = 0, dy/dt = −3
Tangent x = 1; normal y = 0.
04 · The other horizontal tangent
Find the tangent and normal to this curve at t = −1.
Hint
The point is (2, 2) and the tangent is horizontal.
Worked solution
Tangent y = 2; normal x = 2.
05 · Zero parameter speed
For x = t³, y = 2t³ + 1, find the tangent and normal at t = 0.
Hint
Both rates are zero, but elimination gives the straight line y = 2x + 1.
Worked solution
The tangent is y = 2x + 1, and the normal is y = 1 − x/2. Their contact point is (0, 1). Do not infer either line just from 0/0.
05 / Keep trigonometric contact points exact
Use radians and exact values before simplifying.
x = 3 cos t, y = 2 sin t at t = π/3Worked example
Point: (3/2, √3)
Substitute exact cosine and sine values.
dy/dx = −2 cos t/(3 sin t) = −2/(3√3)
Evaluate the rate ratio.
Tangent: y − √3 = −[2/(3√3)](x − 3/2)
The contact coordinates remain exact.
Normal: y − √3 = (3√3/2)(x − 3/2)
The negative reciprocal is positive.
06 · Circle normal
For x = 2 cos t, y = 2 sin t, find the normal at t = π/4.
Hint
The point is (√2, √2). A circle’s normal passes through its centre.
Worked solution
The tangent gradient is −1, so the normal gradient is 1. The normal is y = x, agreeing with the radius through the contact point.
07 · Ellipse at an endpoint
For x = 3 cos t, y = 2 sin t, find the tangent and normal at t = 0.
Hint
The point is (3, 0); dx/dt = 0 and dy/dt = 2.
Worked solution
The tangent is x = 3 and the normal is y = 0.
06 / Find parameters for a given line direction
Translate the line condition into an equation for the gradient.
x = t² + 1, y = t³ − 3t; tangents parallel to y = 3xWorked example
(3t² − 3)/(2t) = 3, with t ≠ 0
Parallel finite lines have equal gradients.
t² − 2t − 1 = 0
Multiply through and simplify.
t = 1 ± √2
Both roots lie in the original unrestricted parameter domain and have nonzero x-rate.
Use each root separately to obtain its point and line
A direction condition can give more than one tangent.
08 · All parallel tangents
For x = t, y = t³, find all tangents parallel to y = 3x + 8.
Hint
The gradient is 3t².
Worked solution
t = ±1. At (1, 1), the tangent is y = 3x − 2. At (−1, −1), it is y = 3x + 2. These are distinct parallel lines.
09 · A perpendicular tangent
For x = t, y = t², find the tangent perpendicular to y = 2x + 1.
Hint
The required tangent gradient is −1/2.
Worked solution
2t = −1/2 gives t = −1/4 and point (−1/4, 1/16). The tangent is y − 1/16 = −(1/2)(x + 1/4), or y = −x/2 − 1/16.
07 / Find a second intersection using a new parameter
The contact parameter is fixed; the other point is unknown.
Tangent to x = t, y = t³ at t = 1Worked example
Contact point (1, 1); tangent y = 3x − 2
First find the tangent in x and y.
At another curve point use x = u, y = u³
A new symbol avoids confusing the unknown point with t = 1.
u³ − 3u + 2 = (u − 1)²(u + 2) = 0
Substitute the parametric coordinates into the line.
u = 1 is the contact; u = −2 gives (−2, −8)
The tangent meets the cubic again.
A tangent describes the local direction. It can intersect a curve elsewhere, and at a point of inflection it can cross the curve at the contact point.
10 · A second tangent intersection
For x = t, y = t³, find the other intersection of the tangent at t = −1 with the curve.
Hint
The tangent is y = 3x + 2.
Worked solution
u³ − 3u − 2 = (u + 1)²(u − 2). The contact is u = −1; the other point is (2, 8).
11 · Apply a parameter restriction
Now restrict the same cubic to t ≥ 0. Does its tangent at t = 1 meet this restricted curve again?
Hint
The second algebraic parameter is −2.
Worked solution
No. The only other root is outside the stated domain. The valid contact parameter u = 1 is a repeated root, not a second distinct point.
08 / Derive a family of tangent equations
Keep the contact parameter fixed while solving for intersections.
x = t, y = t³, with contact parameter aWorked example
Point (a, a³); gradient 3a²
Use a as the fixed contact parameter.
Tangent: y = 3a²x − 2a³
Expand point–slope form.
u³ − 3a²u + 2a³ = (u − a)²(u + 2a)
Substitute an arbitrary point with parameter u.
For a ≠ 0, the other intersection has u = −2a
When a = 0, all three roots coincide at the origin.
12 · An inflection tangent
For x = t, y = t³, describe the tangent and its intersections at t = 0.
Hint
The tangent equation is y = 0.
Worked solution
The tangent is the x-axis. Its only intersection is (0, 0), with root u = 0 of multiplicity three. The curve changes sides of this tangent at the contact point.
09 / Find axis intercepts of a tangent
Substitute zero into the line equation, not the curve.
x = t², y = 2t, at t = 2Worked example
Point (4, 4); dy/dx = 1/t = 1/2
Compute the contact data.
Tangent: y = x/2 + 2
Use point–slope form.
x-intercept (−4, 0); y-intercept (0, 2)
Set y = 0 and x = 0 separately.
Area with the coordinate axes = ½ × 4 × 2 = 4
Use positive lengths even when an intercept is negative.
13 · Normal intercepts
For this curve at t = 2, find the normal’s axis intercepts and the triangle area it forms with the axes.
Hint
The normal has gradient −2 through (4, 4).
Worked solution
The normal is y = −2x + 12. Its intercepts are (6, 0) and (0, 12); the triangle area is 36 square units.
10 / Combine tangent and normal geometry
Draw which triangle the question actually describes.
Tangent and normal at (4, 4) on x = t², y = 2tWorked example
Tangent meets the x-axis at A = (−4, 0)
Use the tangent equation y = x/2 + 2.
Normal meets the x-axis at B = (6, 0)
Use the normal equation y = −2x + 12.
Triangle APB has base 10 and height 4
P is the contact point (4, 4).
Area = ½ × 10 × 4 = 20
This is a different triangle from either line with both axes.
14 · Check the right angle
Show that AP and BP are perpendicular using their direction vectors, with A = (−4, 0), B = (6, 0), P = (4, 4).
Hint
Use AP = (8, 4) and BP = (−2, 4).
Worked solution
The scalar product is 8(−2) + 4(4) = 0, so the vectors are perpendicular.
15 · A second intersection of the normal
The normal at t = 2 on x = t², y = 2t is y = −2x + 12. Find its other intersection with the unrestricted curve.
Hint
Substitute x = u², y = 2u.
Worked solution
2u = −2u² + 12 gives (u − 2)(u + 3) = 0. The other parameter is u = −3, giving (9, −6).
11 / Check the point, direction and domain
A repeated parameter root is not a second distinct point.
- Calculate the contact coordinates from the original equations.
- Find the tangent slope from the ratio of rates.
- Use the negative reciprocal only for finite, nonzero slopes.
- Keep fixed contact parameters separate from unknown intersection parameters.
- Substitute all roots back and enforce the domain.
- For areas, identify the vertices and use positive lengths.
16 · A final consistency check
A claimed normal at (5, 2) has equation 9y = −4x + 37. What is wrong?
Hint
Substitute the contact point.
Worked solution
It gives 18 = 17, so the line misses the point. With this gradient the correct normal is 9y = −4x + 38. Checking the contact point catches an intercept error.
Section 1 of 11 · Start with a point and a direction