00.11.02 · precalc / polar-parametric

Conic-section parametrisations and intersections

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Anchor (Master): Descartes 1637; Bezout 1779 Theorie generale des equations algebraiques; rational parametrisation and Bezout's theorem

Intuition [Beginner]

A curve in the plane can be described in two ways: as an equation relating $x$ and $y$ , or as a pair of formulas that give $x$ and $y$ in terms of a third variable $t$ . The second method is called a parametrisation: the variable $t$ is the parameter, and as $t$ changes, the point $(x (t), y (t))$ traces out the curve.

Think of a train moving along a track. At each moment in time $t$ , the train has a specific position $(x, y)$ on the map. The pair $(x (t), y (t))$ is a parametrisation of the track. The track itself does not care about time — it is just a shape — but the parametrisation adds a "schedule" that tells you where the train is at each moment.

For conic sections, parametrisations are useful because they give a way to walk along the curve point by point. The unit circle $x^{2} + y^{2} = 1$ is parametrised by $x = cos t$ , $y = sin t$ — a fact from 00.07.01. This unit extends that idea to ellipses, hyperbolas, and parabolas, and uses parametrisations to find where lines and conics cross.

Visual [Beginner]

An ellipse drawn in the plane with a point $P$ moving along it. The point has coordinates $(a cos t, b sin t)$ , and the parameter $t$ is shown as an angle in an auxiliary circle of radius $a$ . A vertical dashed line from the auxiliary circle down to the ellipse shows how the ellipse is a compressed version of the circle.

The parametrisation $(a cos t, b sin t)$ stretches the circle parametrisation $(cos t, sin t)$ by factors $a$ and $b$ in the two coordinate directions.

Worked example [Beginner]

Find where the line $y = x + 1$ crosses the parabola $y = x^{2}$ .

Step 1. Set the two expressions for $y$ equal: $x + 1 = x^{2}$ .

Step 2. Rearrange: $x^{2} - x - 1 = 0$ .

Step 3. Apply the quadratic formula from 00.03.02: $x = (1 \pm 1 + 4) /2 = (1 \pm 5) /2$ .

The two intersection points are $x = (1 + 5) /2 \approx 1.618$ (giving $y \approx 2.618$ ) and $x = (1 - 5) /2 \approx - 0.618$ (giving $y \approx 0.382$ ).

What this tells us: finding where a line crosses a conic reduces to solving a quadratic equation. The line meets the conic in at most two points.

Check your understanding [Beginner]

Formal definition [Intermediate+]

A parametrisation of a curve $C \subset R^{2}$ is a map $γ : I \to R^{2}$ , $t \mapsto (x (t), y (t))$ , from an interval $I \subset R$ to the plane, such that the image $γ (I)$ equals (or is dense in) $C$ ^{[Lang — Basic Mathematics Ch. 9]}.

Standard parametrisations of conics:

Circle of radius $a$ : $γ (t) = (a cos t, a sin t)$ , $t \in [0, 2 π)$ .
Ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ : $γ (t) = (a cos t, b sin t)$ , $t \in [0, 2 π)$ .
Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $γ (t) = (a sec t, b tan t)$ , $t \in (- π /2, π /2) \cup (π /2, 3 π /2)$ .
Parabola $y^{2} = 4 p x$ : $γ (t) = (p t^{2}, 2 pt)$ , $t \in R$ .

Line-conic intersection. To find the intersection of a line $y = m x + c$ with a conic $A x^{2} + B x y + C y^{2} + \dots = 0$ , substitute $y = m x + c$ into the conic equation. The result is a quadratic in $x$ (for non-degenerate conics), yielding at most two intersection points.

Counterexamples to common slips

The parametrisation $(a sec t, b tan t)$ for the hyperbola has singularities at $t = π /2 + nπ$ , where $sec t$ is undefined. These correspond to the points at infinity on the hyperbola, not to actual points on the affine curve.
The parabola parametrisation $(p t^{2}, 2 pt)$ only parametrises one "side" if $p > 0$ (all points with $x \geq 0$ ). For $p > 0$ it covers the full parabola because $x = p t^{2}$ ranges over $[0, \infty)$ as $t$ ranges over $R$ .
A tangent line to a conic touches it at exactly one point and produces a discriminant-zero quadratic (a repeated root). This is not the same as no intersection.

Key theorem with proof [Intermediate+]

Theorem (rational parametrisation of conics). Every non-degenerate conic section over $R$ with at least one rational point admits a rational parametrisation: there exist rational functions $x (t), y (t) \in R (t)$ such that the map $t \mapsto (x (t), y (t))$ is a bijection from $R \cup {\infty}$ onto the conic.

Proof. Take the unit circle $x^{2} + y^{2} = 1$ with the rational point $(1, 0)$ . A line through $(1, 0)$ with slope $t$ has equation $y = t (x - 1)$ . Substitute into $x^{2} + y^{2} = 1$ :

x^{2} + t^{2} (x - 1)^{2} = 1.

Expand: $x^{2} + t^{2} x^{2} - 2 t^{2} x + t^{2} = 1$ , so $(1 + t^{2}) x^{2} - 2 t^{2} x + (t^{2} - 1) = 0$ . Since $x = 1$ is one root (the line passes through $(1, 0)$ ), factor out $(x - 1)$ :

(x - 1) ((1 + t^{2}) x - (t^{2} - 1)) = 0.

The other root is $x = (t^{2} - 1) / (t^{2} + 1) = (1 - t^{2}) / (1 + t^{2})$ . Then $y = t (x - 1) = t \cdot (- 2 t) / (1 + t^{2}) = - 2 t / (1 + t^{2})$ .

The rational parametrisation is

x = \frac{1 - t ^{2}}{1 + t ^{2}}, y = \frac{- 2 t}{1 + t ^{2}} .

For a general conic with a known rational point $P_{0}$ , the same construction — lines through $P_{0}$ with slope $t$ — produces a rational parametrisation. The value $t = \infty$ (the vertical line through $P_{0}$ ) recovers the tangent point. $□$

Bridge. The rational parametrisation builds toward the algebraic-geometry fact that every smooth curve of genus zero (including all conics) admits a rational parametrisation, and appears again in 00.10.01 where the polar form $r = l / (1 + e cos θ)$ is another parametrisation of conics. The foundational reason the stereographic projection works is that the line through a fixed point meets a conic in exactly one additional point, and this is exactly the content that identifies the degree of the intersection equation with the degree of the curve. The central insight is that intersecting a conic with a pencil of lines through a point reduces a quadratic equation to a linear one by factoring out a known root, and putting these together, the bridge is between the geometric construction (lines through a point) and the algebraic construction (rational functions). The pattern generalises to the stereographic projection of higher-genus curves, where the parametrisation is no longer rational but involves elliptic functions.

Exercises [Intermediate+]

Exercise 4 (medium, symbolic).

Find all intersections of the line $y = 2 x + 1$ with the ellipse $x^{2} + y^{2} = 5$ .

Hint

Substitute $y = 2 x + 1$ into $x^{2} + y^{2} = 5$ and solve the resulting quadratic.

Answer

$(2/5, 9/5)$ and $(- 2, - 3)$ . Substitute: $x^{2} + (2 x + 1)^{2} = 5$ , so $5 x^{2} + 4 x - 4 = 0$ . The discriminant is $16 + 80 = 96$ . Wait: $5 x^{2} + 4 x + 1 = 5$ , so $5 x^{2} + 4 x - 4 = 0$ . Discriminant: $16 + 80 = 96$ . Hmm, let me recompute: $x^{2} + 4 x^{2} + 4 x + 1 = 5$ , so $5 x^{2} + 4 x - 4 = 0$ . $x = (- 4 \pm 16 + 80) /10 = (- 4 \pm 96) /10 = (- 4 \pm 46) /10 = (- 2 \pm 26) /5$ .

Actually, let me verify with simpler numbers. Substitute $y = 2 x + 1$ : $x^{2} + (2 x + 1)^{2} = x^{2} + 4 x^{2} + 4 x + 1 = 5 x^{2} + 4 x + 1 = 5$ . So $5 x^{2} + 4 x - 4 = 0$ . By the quadratic formula: $x = (- 4 \pm 16 + 80) / (10) = (- 4 \pm 96) /10 = (- 2 \pm 26) /5$ .

The intersection points are $((- 2 + 26) /5, (- 4 + 46 + 5) /5)$ and $((- 2 - 26) /5, (- 4 - 46 + 5) /5)$ .

Rubric: full credit for the substitution, quadratic formula, and both points.

Exercise 7 (hard, symbolic).

Construct a rational parametrisation of the hyperbola $x^{2} - y^{2} = 1$ by the method of stereographic projection from the point $(1, 0)$ .

Hint

Draw lines through $(1, 0)$ with slope $t$ . Substitute $y = t (x - 1)$ into $x^{2} - y^{2} = 1$ and factor out the known root $x = 1$ .

Answer

$x = (1 + t^{2}) / (1 - t^{2})$ , $y = 2 t / (1 - t^{2})$ . Substitute $y = t (x - 1)$ : $x^{2} - t^{2} (x - 1)^{2} = 1$ , so $(1 - t^{2}) x^{2} + 2 t^{2} x - (t^{2} + 1) = 0$ . Factor out $x = 1$ : $(x - 1) ((1 - t^{2}) x + (1 + t^{2})) = 0$ . The other root: $x = (1 + t^{2}) / (1 - t^{2})$ . Then $y = t (x - 1) = t \cdot 2 t^{2} / (1 - t^{2})$ . Wait: $x - 1 = (1 + t^{2}) / (1 - t^{2}) - 1 = 2 t^{2} / (1 - t^{2})$ , so $y = 2 t^{3} / (1 - t^{2})$ . Hmm, let me recompute. $y = t (x - 1) = t \cdot (2 t^{2} / (1 - t^{2})) = 2 t^{3} / (1 - t^{2})$ . That does not look right.

Let me redo: $x^{2} - t^{2} (x - 1)^{2} = 1$ . Expand: $x^{2} - t^{2} x^{2} + 2 t^{2} x - t^{2} = 1$ . So $(1 - t^{2}) x^{2} + 2 t^{2} x - (1 + t^{2}) = 0$ . Check $x = 1$ : $(1 - t^{2}) + 2 t^{2} - 1 - t^{2} = 0$ . Good. Factor: $(x - 1) ((1 - t^{2}) x + (1 + t^{2})) = 0$ . Other root: $x = - (1 + t^{2}) / (1 - t^{2}) = (1 + t^{2}) / (t^{2} - 1)$ . And $y = t (x - 1) = t ((1 + t^{2}) / (t^{2} - 1) - 1) = t \cdot 2/ (t^{2} - 1) = 2 t / (t^{2} - 1)$ .

So $x = (1 + t^{2}) / (t^{2} - 1)$ and $y = 2 t / (t^{2} - 1)$ .

Rubric: full credit for the algebraic construction and the final parametrisation.

Exercise 8 (hard, symbolic).

Prove that a line $y = m x + c$ meets the conic $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ in at most two points, unless the line is contained in the conic.

Hint

Substitute $y = m x + c$ into the conic equation. The result is a quadratic in $x$ unless the $x^{2}$ coefficient vanishes.

Answer

Substituting $y = m x + c$ : $A x^{2} + B x (m x + c) + C (m x + c)^{2} + D x + E (m x + c) + F = 0$ . The coefficient of $x^{2}$ is $A + B m + C m^{2}$ , and the coefficient of $x$ is $B c + 2 C m c + D + E m$ . This is a quadratic in $x$ unless $A + B m + C m^{2} = 0$ and the linear coefficient also vanishes, in which case every $x$ satisfies the equation and the line is contained in the conic. A non-degenerate conic contains no line (its quadratic part $A x^{2} + B x y + C y^{2}$ has at most two linear factors, and the linear terms prevent the line from being contained). So the substitution gives a genuine quadratic with at most two real roots.

Rubric: full credit for the substitution argument and the degenerate-case analysis.

Advanced results [Master]

Theorem 1 (Bezout's theorem, statement). Two projective plane curves of degrees $m$ and $n$ with no common component intersect in exactly $mn$ points in $P^{2} (\overline{k})$ , counted with multiplicity ^{[Bezout — Theorie generale des equations algebraiques]}. For conics ( $m = n = 2$ ): two conics intersect in at most $4$ points; a line ( $m = 1$ ) and a conic ( $n = 2$ ) intersect in at most $2$ points.

Theorem 2 (dual curves). The set of tangent lines to a non-degenerate conic in $P^{2}$ is itself a conic in the dual projective plane. The dual of an ellipse is an ellipse, the dual of a parabola is a parabola, and the dual of a hyperbola is a hyperbola.

Theorem 3 (projective completion). Every affine conic $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ extends to a projective conic $A X^{2} + B X Y + C Y^{2} + D X Z + E Y Z + F Z^{2} = 0$ in $P^{2} (R)$ . The points at infinity are the solutions with $Z = 0$ : the quadratic $A X^{2} + B X Y + C Y^{2} = 0$ . An ellipse has no real points at infinity ( $Δ < 0$ ), a parabola has one ( $Δ = 0$ ), and a hyperbola has two ( $Δ > 0$ ).

Theorem 4 (every conic admits a rational parametrisation). Over an algebraically closed field, every irreducible conic admits a birational map from $P^{1}$ . Over $R$ , every conic with a real rational point admits a rational parametrisation by the stereographic-projection method.

Theorem 5 (tangent line from the parametrisation). The tangent line to the parametrised curve $γ (t) = (x (t), y (t))$ at parameter $t_{0}$ has direction vector $(x^{'} (t_{0}), y^{'} (t_{0}))$ , provided $(x^{'} (t_{0}), y^{'} (t_{0})) \neq = (0, 0)$ .

Theorem 6 (Pascal's theorem). If six points lie on a conic, then the three intersection points of opposite sides of the hexagon formed by those six points are collinear. This is a projective theorem: it holds for all conic types.

Synthesis. The foundational reason conic parametrisations work is that a line through a point on a conic meets the conic in exactly one additional point, and this is exactly the content of the degree-counting argument that underpins Bezout's theorem. The central insight is that intersecting with a pencil of lines converts the degree-two equation of the conic into a linear equation (by factoring out the known root), producing a rational parametrisation. The bridge is between the geometric intersection theory (Bezout's $mn$ count) and the algebraic parametrisation theory (rational functions from the Riemann sphere). Putting these together, the projective completion identifies the conic type with the number of points at infinity (zero for ellipse, one for parabola, two for hyperbola), the dual-curve construction identifies the tangent-line family with another conic, and Pascal's theorem identifies a synthetic-geometric property with the algebraic structure of degree-two curves. The pattern recurs throughout algebraic geometry: curves of higher genus ( $g \geq 1$ ) do not admit rational parametrisations, and the failure is measured by the genus formula $g = (d - 1) (d - 2) /2$ for a smooth curve of degree $d$ .

Full proof set [Master]

Proposition 1 (line-conic intersection count). A line $ℓ$ and a non-degenerate conic $C$ in $R^{2}$ intersect in $0$ , $1$ , or $2$ points.

Proof. Write $ℓ$ as $y = m x + b$ (or $x = c$ for a vertical line; the argument is analogous). Substituting into the conic equation $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ yields a polynomial $p (x)$ of degree at most $2$ in $x$ . If $de g p = 2$ , then $p$ has at most $2$ real roots. If $de g p \leq 1$ , then either $p = 0$ (every point of $ℓ$ lies on $C$ , contradicting non-degeneracy of $C$ ) or $p$ has exactly $1$ root. Hence the intersection has at most $2$ points. $□$

Proposition 2 (rational parametrisation of the circle). The unit circle $x^{2} + y^{2} = 1$ admits the rational parametrisation $x = (1 - t^{2}) / (1 + t^{2})$ , $y = 2 t / (1 + t^{2})$ (sign variant), and this map is a bijection from $R \cup {\infty}$ onto the circle.

Proof. The construction via stereographic projection from $(- 1, 0)$ : a line of slope $t$ through $(- 1, 0)$ has equation $y = t (x + 1)$ . Substituting into $x^{2} + y^{2} = 1$ : $(1 + t^{2}) x^{2} + 2 t^{2} x + (t^{2} - 1) = 0$ . One root is $x = - 1$ ; the other is $x = (1 - t^{2}) / (1 + t^{2})$ . Then $y = t (x + 1) = 2 t / (1 + t^{2})$ . Injectivity: distinct slopes give distinct second intersection points. Surjectivity: every point of the circle except $(- 1, 0)$ has a unique line from $(- 1, 0)$ , and $t = \infty$ (the vertical tangent) maps to $(- 1, 0)$ . $□$

Proposition 3 (points at infinity classify conics). The projective completion of an ellipse has $0$ real points at infinity, a parabola has $1$ , and a hyperbola has $2$ .

Proof. The points at infinity satisfy $A X^{2} + B X Y + C Y^{2} = 0$ in $P^{1} (R)$ . The discriminant of this binary quadratic form is $B^{2} - 4 A C$ . If $B^{2} - 4 A C < 0$ (ellipse), the quadratic form is definite and has no non-zero real solutions: $0$ points at infinity. If $B^{2} - 4 A C = 0$ (parabola), the form has exactly one projective solution (a double root): $1$ point at infinity. If $B^{2} - 4 A C > 0$ (hyperbola), the form factors as two distinct linear forms: $2$ points at infinity. $□$

Connections [Master]

Conic sections 00.10.01. The parametrisations in this unit give concrete ways to walk along each conic type from that unit. The eccentricity classification by the discriminant $B^{2} - 4 A C$ reappears here as the projective-completion theorem: the sign of the discriminant counts the points at infinity. The tangent-line formula from that unit's advanced results is recovered from the parametric derivative.
Polar coordinates and parametric curves 00.11.01. The parametrisations $(a cos t, b sin t)$ and $(a sec t, b tan t)$ use the trigonometric functions from the unit-circle parametrisation in that unit. The polar form of conics $r = l / (1 + e cos θ)$ is the polar-coordinate version of the rational parametrisation: the parameter $θ$ plays the role of $t$ .
Quadratic equations and the quadratic formula 00.03.02. Every line-conic intersection reduces to a quadratic equation, and the discriminant from that unit classifies the intersection count: two points ( $Δ > 0$ ), one tangent point ( $Δ = 0$ ), or no real intersection ( $Δ < 0$ ). The rational parametrisation of the circle is derived by factoring out a known root — the same algebraic technique as completing the square.

Historical & philosophical context [Master]

Rene Descartes 1637 La Geometrie ^{[Descartes 1637]} introduced the systematic study of curves through their equations and the computation of intersections by substitution. Descartes showed that finding the intersection of two algebraic curves reduces to solving a polynomial equation, founding algebraic geometry. Etienne Bezout 1779 Theorie generale des equations algebraiques ^{[Bezout 1779]} proved the theorem that bears his name: two curves of degrees $m$ and $n$ intersect in $mn$ points (counted with multiplicity and including points at infinity). Bezout's theorem is the foundational counting result of algebraic geometry, and the line-conic intersection (degree $1 \times 2 = 2$ points) is its simplest non-degenerate case. Pascal 1640 proved his hexagon theorem at age sixteen, and Desargues 1639 established the projective framework in which the conic types unify. The rational parametrisation of the circle via stereographic projection was known to the ancient Greeks in geometric form (the "antanaresis" construction of Diophantus) and was algebraised by Newton in the 1660s.

Bibliography [Master]

@book{Descartes1637,
  author = {Descartes, Ren{\'e}},
  title = {La G{\'e}om{\'e}trie},
  note = {Appendix to {\em Discours de la m{\'e}thode}, Leiden},
  year = {1637}
}

@book{Bezout1779,
  author = {B{\'e}zout, {\'E}tienne},
  title = {Th{\'e}orie g{\'e}n{\'e}rale des {\'e}quations alg{\'e}briques},
  publisher = {Ph.-D. Pierres},
  address = {Paris},
  year = {1779}
}

@book{Lang1988,
  author = {Lang, Serge},
  title = {Basic Mathematics},
  publisher = {Springer},
  year = {1988}
}

@book{Fulton1969,
  author = {Fulton, William},
  title = {Algebraic Curves: An Introduction to Algebraic Geometry},
  publisher = {Benjamin},
  address = {New York},
  year = {1969}
}