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Tangents and chords of parametric curves

Find chord equations from parametric points and use a quadratic discriminant to find tangents. Explore parallel lines, intersection counts and cases where repeated parameter roots need care.

Before you startParametric intersections, straight-line equations and the discriminant

01 / Distinguish a chord from a tangent

A chord joins two distinct points; a tangent describes the local direction.

Find the coordinates of the relevant points before using straight-line geometry. For some curves, substituting a line into the parametric rules gives a quadratic. A repeated root can then identify a tangent, provided the parameter describes the geometry faithfully.

Coordinates first; then line geometry or a justified discriminant test.

In the model x = t, y = t², each parameter gives a different point on the familiar parabola y = x². Move a line parallel to itself to compare two crossings, tangency and no meeting.

Move a parallel lineExplore

Gradient mOffset above tangent

x = t; y = t², for all real t.

Line y = 0; t² = 0.

Discriminant = 0.

One point: (0, 0). The line is tangent here.

The offset is measured above the parallel tangent, so c = −m²/4 + offset. Blue shows the parabola; gold marks all intersections. The picture displays −2.5 ≤ t ≤ 2.5.

02 / Find the equation of a chord

Use two distinct coordinate pairs.

For x = t + 1, y = t², find the chord joining t = −1 and t = 2.Worked example

P = (0, 1), Q = (3, 4)

Substitute each parameter into both rules.

Gradient = (4 − 1)/(3 − 0) = 1

Use changes in y and x.

y − 1 = x − 0, so y = x + 1

Check both endpoints in the final equation.

If the x-coordinates are equal, the chord is vertical: write x = constant. Do not divide by a zero horizontal change.

03 / Derive a chord equation in terms of its parameters

Factor before cancelling, and state the restriction.

For x = t, y = t², let P use t = p and Q use t = q, with p ≠ q.Worked example

P = (p, p²), Q = (q, q²)

Different parameters give distinct points for this curve.

m = (q² − p²)/(q − p) = p + q

The cancellation uses p ≠ q.

y − p² = (p + q)(x − p)

Point-gradient form.

y = (p + q)x − pq

A useful chord equation.

Putting q = p into the final expression suggests the tangent y = 2px − p². This is a limit of chords, not permission to divide by q − p when it is zero.

Watch a chord approach the tangent

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Keep vertical chords in the geometry

A gradient formula can be undefined while the line is perfectly valid.

For x = t², y = t, join parameters p and q, p ≠ q.Worked example

Points (p², p), (q², q)

Begin with both pairs.

x = (p + q)y − pq

This line equation works even when p + q = 0.

If q = −p, then x = p²

The chord is vertical for p ≠ 0.

For p = −2, q = 2 the chord is x = 4. Both endpoints, (4, −2) and (4, 2), lie on it.

05 / Find a tangent using a quadratic

A double intersection root is a useful test in this parabola example.

Find the line of gradient 4 tangent to x = t, y = t².Worked example

Write y = 4x + c

The gradient is fixed but the intercept is unknown.

t² − 4t − c = 0

Substitute the parameter rules.

Discriminant = 16 + 4c = 0

For this parabola, tangency means one repeated intersection.

c = −4, t = 2

The point of contact is (2, 4).

Tangent: y = 4x − 4

Check the contact point in both equations.

The leading coefficient must really be non-zero before a quadratic discriminant is used. If an unknown value makes it vanish, examine the resulting linear or constant equation separately.

06 / Classify all parallel lines

For a fixed gradient, the intercept controls how often the line meets the curve.

Classify y = mx + c against x = t, y = t².Worked example

t² − mt − c = 0

The leading coefficient is always 1.

D = m² + 4c

Count distinct real roots.

c > −m²/4: two intersections

The line is above the parallel tangent.

c = −m²/4: tangent at (m/2, m²/4)

One repeated root.

c < −m²/4: no intersections

The line lies below the tangent.

These statements concern the whole parabola. A restricted parameter interval can remove intersections and produce a single remaining crossing that is not a tangent.

07 / Find parallel tangents to a circle

Elimination can make the discriminant test easier.

Find the tangents parallel to y = x for x = 2 cos θ, y = 2 sin θ, over a complete turn.Worked example

x² + y² = 4

The rules trace the whole circle.

Substitute y = x + c: 2x² + 2cx + c² − 4 = 0

Now solve for intersection x-coordinates.

D = 4c² − 8(c² − 4) = 32 − 4c²

Tangency requires D = 0.

c = ±2√2

Two parallel tangents.

Contacts: (−√2, √2) and (√2, −√2)

Use x = −c/2 and y = c/2.

With a restricted arc, verify that the calculated point is actually reached by an allowed angle. A tangent to the full circle need not touch the selected arc.

08 / Check the contact point and the meaning of one root

One surviving intersection is not always a tangent.

For x = t, y = t² with t ≥ 0, the line y = 1 has just one permitted meeting, (1, 1). It crosses the parabola there. The missing point (−1, 1) was removed by the domain; no tangency was created.

For x = t², y = t, is x = 0 tangent at the origin?Worked example

x = y² is the Cartesian curve

It opens to the right.

The line x = 0 meets only at (0, 0)

The tangent is vertical.

The chord between t = −h and t = h is x = h²

As h tends to 0, these vertical chords approach x = 0.

A gradient-intercept form y = mx + c cannot describe a vertical tangent. Keep the line form appropriate to the geometry.

09 / Do not trust root multiplicity alone

The parameter itself can create a repeated algebraic root.

Take x = t³, y = t⁶ and intersect with y = x.Worked example

t⁶ = t³, so t³(t³ − 1) = 0

The parameter root t = 0 has multiplicity three.

Eliminate t: y = x²

The curve is still an ordinary parabola.

The line y = x crosses at the origin and at (1, 1)

A repeated parameter root at the origin did not make this line tangent.

The line y = −2x − 1 touches x = t, y = t². Verify the contact by substitution.

Hint

Rearrange t² = −2t − 1.

Worked solution

(t + 1)² = 0, so t = −1 gives (−1, 1). It lies on the line; the repeated intersection agrees with the parabola tangent.

11 / Recap

Use the algebra with its geometric conditions.

A chord needs two distinct points; calculate both coordinates first.
Vertical lines require a separate gradient case or a line form that keeps it.
A quadratic discriminant can locate tangency when the intersection geometry justifies it.
Find and check the contact point, including the parameter domain.
A restricted domain or a repeated parameter value can mislead a root-count shortcut.

Section 1 of 11 · Distinguish a chord from a tangent