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Parametric curve intersections

Find where parametric curves meet axes, lines, circles and other curves. Substitute both coordinate rules, solve for the parameter, filter its domain and count distinct intersection points.

Before you startParametric coordinates, polynomial and trigonometric equations

01 / Substitute both coordinates into the other equation

A shared point must satisfy both descriptions.

If a curve is given by x = p(t), y = q(t), substitute those expressions into the line or curve it meets. Solve the resulting equation for t, keep valid parameters, and then calculate the coordinate pairs.

Substitute → solve for t → filter → find (x, y) → remove duplicate points

The model makes the domain filter visible. A real algebraic root may lie outside the permitted part of the curve.

Find the shared pointsExplore

Line gradient mLine intercept cParameter domain

x = t²; y = t.

Line y = x; parameter equation t² − t = 0.

Candidate parameters: 0 and 1.

2 distinct intersections: (0, 0) and (1, 1).

The blue curve keeps only the selected domain. Gold points satisfy both rules. Here y = t makes different valid parameters different points; other parameterisations can revisit a point.

02 / Intersect a curve with a line

Do not equate a parameter directly to a coordinate.

Find the intersections of x = t², y = t + 1 with y = x + 1, for t real.Worked example

t + 1 = t² + 1

Substitute both x and y into the line.

t² − t = t(t − 1) = 0

Rearrange without dividing by t.

t = 0 or t = 1

Keep both roots.

Points (0, 1) and (1, 2)

Use the original coordinate rules, then check the line.

Dividing by t would lose the first intersection. Factor an equation equal to zero instead.

Watch the line and shared points move

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

03 / Treat each axis as a simple equation

The x-axis has y = 0; the y-axis has x = 0.

For x = t² − 4, y = t + 1, find the axis crossings.Worked example

x-axis: t + 1 = 0, so t = −1

The point is (−3, 0).

y-axis: t² − 4 = 0, so t = −2 or 2

The points are (0, −1) and (0, 3).

A vertical line such as x = 5 is handled through the x-rule. There is no need to put that line into a gradient-intercept form.

04 / Filter roots before reporting points

The parameter domain can remove an otherwise valid intersection.

Find where x = t², y = t meets y = 2 − x, with 0 ≤ t ≤ 2.Worked example

t = 2 − t²

Substitute into the line.

(t + 2)(t − 1) = 0

Candidate parameters are −2 and 1.

Reject t = −2

It is outside 0 ≤ t ≤ 2.

The only intersection is (1, 1)

The negative candidate would have been (4, −2) on the unrestricted parabola.

Test original denominator restrictions too. Multiplying an equation by a denominator does not make a pole valid.

05 / Substitute into a circle equation

A squared parameter may reduce a quartic to a quadratic.

Intersect x = t, y = t² with x² + y² = 2, for t real.Worked example

t² + t⁴ = 2

Substitute x and y.

Let u = t²: u² + u − 2 = 0

This is a quadratic in u, with u ≥ 0.

(u + 2)(u − 1) = 0, so u = 1

Reject u = −2 because t is real.

t = ±1, giving (−1, 1) and (1, 1)

Both coordinates satisfy the circle.

The temporary variable u is not the parameter or a coordinate. Convert all the way back to t and then to points.

06 / Use the same method for any Cartesian curve

Simplify carefully and retain every root.

Find intersections of x = t, y = t² − 1 with y² = x² + 1.Worked example

(t² − 1)² = t² + 1

Substitute before expanding.

t⁴ − 3t² = 0

Collect terms.

t²(t² − 3) = 0

Parameters are 0, √3 and −√3.

Points (0, −1), (√3, 2), (−√3, 2)

The repeated root at t = 0 still gives just one point.

An algebraic repeated root does not mean that two separate intersections occur at the same coordinates. Count distinct positions.

07 / Solve the complete trigonometric parameter equation

Keep every solution in the given interval.

Intersect x = 2 cos θ, y = 2 sin θ with y = x, for 0 ≤ θ < 2π.Worked example

sin θ = cos θ

Substitute and cancel the common factor 2.

θ = π/4 or 5π/4

Both satisfy the original equation in the interval.

(√2, √2) and (−√2, −√2)

Use both coordinate rules.

If dividing by a trigonometric expression, first check whether its zeros solve the original equation. You may instead rearrange or use an identity that preserves all roots.

08 / Count points separately from parameter roots

A curve may reach one intersection more than once.

Intersect x = t² − 1, y = t³ − t with the x-axis for −2 ≤ t ≤ 2.Worked example

t(t − 1)(t + 1) = 0

Parameters are −1, 0 and 1.

t = −1 and t = 1 both give (0, 0)

Two valid parameters produce one point.

t = 0 gives (−1, 0)

There are two distinct intersections in total.

Do not remove a parameter before checking coordinates: multiple roots can correspond to different points, and different roots can correspond to the same point.

09 / Use a known point to determine an unknown constant

The point supplies two coordinate equations sharing one parameter.

The curve x = 2t + a, y = t² + 1 passes through (7, 5). Find the possible values of a.Worked example

t² + 1 = 5 gives t = ±2

Use the coordinate rule without the unknown constant first.

If t = 2, 7 = 4 + a, so a = 3

One possibility.

If t = −2, 7 = −4 + a, so a = 11

A second possibility.

If the question also states t ≥ 0, only a = 3 remains

The domain can resolve the ambiguity.

A known point does not always determine one unique constant. Report every valid possibility unless extra information selects one.

10 / Verify candidates in the original equations

The parameter equation is a tool, not the final answer.

For each candidate, check that the parameter is defined and permitted, calculate the original (x, y), and substitute that point into the other equation. Keep exact values where possible until the final requested rounding.

Could x = 1/(t − 1), y = t + 2 meet the line x = 0?Worked example

1/(t − 1) = 0 has no solution

A non-zero numerator cannot produce zero for a finite valid denominator.

The limiting x-value 0 is not reached

An asymptote is not an intersection.

For x = t², y = t, the line y = x gives t² = t. A learner divides by t and reports only (1, 1). Correct this.

Hint

Bring everything to one side and factor.

Worked solution

t(t − 1) = 0, so t = 0 or 1. The missing point is (0, 0); both it and (1, 1) are intersections.

12 / Recap

Solve for valid parameters, then report distinct shared points.

Substitute both original coordinate rules into the other equation.
Keep every algebraic or trigonometric root until it has been checked.
Reject excluded parameters and poles.
Calculate (x, y) and verify the other equation.
Count distinct coordinate pairs, not root multiplicity or repeat visits.

Section 1 of 12 · Substitute both coordinates into the other equation