Conic sections and coordinate geometry

Prepare to connect distances and equations

You will use signed coordinates, the Pythagorean distance formula and completing the square. Revisit the preparation links if plotting a point, finding a distance or rewriting x² + 8x as (x + 4)² − 16 is unfamiliar.

On a first reading, follow the four distance definitions, their adjacent diagrams and the feature tables, then try the worked examples. Read the first paragraph of each ellipse and hyperbola proof to learn its defining condition; you can return to the remaining algebraic paragraphs afterward. The eccentricity classification and complete parameterisations are a return path once you can read the standard equations.

A conic is a complete set of points, not a handful of marks

Fix O = (−4, 3) and ask for every point exactly 5 units away. P = (−1, 7) qualifies because its horizontal and vertical displacements are 3 and 4, and √(3² + 4²) = 5. The centre O does not qualify: its distance from itself is zero. The qualifying points form a circle, including all positions around O rather than only the marked point.

A locus is the set of every point satisfying a stated condition. Conic sections can arise from plane sections of a double cone; here we study their plane distance definitions and coordinate equations. Our ordinary circles, ellipses, hyperbolas and parabolas are nondegenerate curves. Some boundary data instead describe a point, segment, line, pair of rays or empty set, and a formula with a zero denominator is not a defined equation.

Use a Euclidean coordinate plane with perpendicular axes and the same length unit on both axes. The distance from P = (x, y) to F = (f, g) is √((x − f)² + (y − g)²). The distance from P to a vertical line x = d is |x − d|, and to a horizontal line y = d is |y − d|. These line distances are perpendicular distances; a chosen sloping segment to another point on the line is generally longer.

For a translated centre or vertex (h, k), write X = x − h and Y = y − k. Translation preserves all lengths and perpendicular directions. This lesson's standard forms have their principal or symmetry axes parallel to the coordinate axes. A general quadratic containing an xy term can require rotated axes; it must not be forced into the axis-aligned forms merely by ignoring that term.

To prove that a distance definition and an equation describe the same locus, prove both implications. Squaring nonnegative distances preserves an equality, but after isolating or squaring a signed expression we must check its sign before reversing the calculation. An exact test of a few points demonstrates membership for those points; it does not establish the full locus.

A circle has one positive radius and a fixed centre

A circle of centre (h, k) and radius r > 0 consists exactly of the points with √(X² + Y²) = r. Both sides are nonnegative, so this is equivalent to X² + Y² = r². Conversely, the squared equation implies the original distance equals the positive root r. The four coordinate-direction endpoints are (h ± r, k) and (h, k ± r), and the full curve is symmetric about every line through its centre.

For an equation x² + y² + Dx + Ey + F = 0, completing both squares gives (x + D/2)² + (y + E/2)² = (D² + E²)/4 − F. Thus the candidate centre is (−D/2, −E/2). Call the right side R. A positive R gives an ordinary circle of radius √R, R = 0 gives only the centre point, and R < 0 gives no real point because a sum of squares cannot be negative.

The input 'radius −2' is outside the original distance definition: a nonnegative distance cannot equal −2. Squaring that input to obtain 4 would replace the original impossible distance condition with a different one. A supplied radius zero likewise gives a point rather than an ordinary positive-radius circle; retain the distinction instead of drawing an arbitrary positive circle.

Equal coefficients of x² and y² can identify a circle after dividing by their common nonzero coefficient and completing squares, provided no xy term remains. The resulting right side still needs its sign check. The centre of a positive-radius circle is not on its boundary, and points inside its disk should not be confused with points on the circle.

An ellipse keeps the sum of two focus distances constant

Place two distinct foci at (h − c, k) and (h + c, k), with c > 0. For P, let d₋ = √((X + c)² + Y²) and d₊ = √((X − c)² + Y²). The noncircular ellipse is defined by d₋ + d₊ = 2a with a > c. The sum must strictly exceed the focus separation 2c: equality would place P on the segment between the foci rather than on an ordinary ellipse.

The squared-distance difference is d₋² − d₊² = 4cX. If their sum is 2a, their difference is 2cX/a, so d₋ = a + cX/a and d₊ = a − cX/a. Substituting either squared distance and simplifying gives X²/a² + Y²/(a² − c²) = 1. Put b² = a² − c² > 0. Then a > b > 0 and the standard horizontal equation is X²/a² + Y²/b² = 1.

The reverse implication needs the signs. The standard equation implies |X| ≤ a. Since c < a, both a + cX/a and a − cX/a are strictly positive. Substituting Y² = b²(1 − X²/a²) into the two distance squares gives d₋² = (a + cX/a)² and d₊² = (a − cX/a)². Taking their nonnegative square roots therefore gives d₋ + d₊ = 2a. This proves the entire locus in both directions, without assuming that a sketch settles the unsquaring step.

The relation c² = a² − b² places each focus inside the corresponding vertex: 0 < c < a. If the two foci coincide, c = 0, their distance sum 2a with a > 0 reduces to distance a from the common focus. The equation then has a = b and is a circle, the circular special case of an ellipse. There is no distinguished longer axis in that equal-axis case.

The larger ellipse denominator marks the major-axis direction

For a noncircular ellipse, a denotes the positive major semiaxis and b the positive minor semiaxis, so a > b > 0. In X²/a² + Y²/b² = 1, the major axis is horizontal; in X²/b² + Y²/a² = 1, it is vertical. The denominators are squares of semiaxis lengths, not the lengths themselves. The full major and minor axis lengths are 2a and 2b.

For the horizontal form, the vertices are (h ± a, k), the minor-axis endpoints are (h, k ± b), and the foci are (h ± c, k) where c = √(a² − b²) > 0. For the vertical form, exchange the X and Y roles: vertices (h, k ± a), minor endpoints (h ± b, k), and foci (h, k ± c). Each axis of symmetry passes through the centre (h, k).

Setting one translated coordinate to zero finds the corresponding axis endpoints. The full horizontal locus has h − a ≤ x ≤ h + a and k − b ≤ y ≤ k + b; each bound is attained. These statements follow from the two nonnegative terms in the equation and the actual endpoints. On a vertical ellipse the bounds instead use b horizontally and a vertically.

Do not declare X²/4 + Y²/9 = 1 invalid merely because the denominator under X is smaller. It is a valid vertical ellipse, with a = 3 and b = 2. Assign the symbols from the major-axis direction after reading the equation. A centre shown at (h, k) and two foci shown inside the curve are reference features, not additional points of the boundary.

A hyperbola retains both signs of a distance difference

Use distinct foci (h − c, k) and (h + c, k) with c > 0, and the same distance names d₋ and d₊. A nondegenerate hyperbola is the locus |d₋ − d₊| = 2a with 0 < a < c. The absolute value includes both branches. The equation d₋ − d₊ = 2a by itself selects the right branch, while the opposite signed difference selects the left branch.

Let δ = d₋ − d₊. Since δ(d₋ + d₊) = 4cX and d₋ + d₊ > 0, δ has the same sign as X. Under |δ| = 2a, it is nonzero. Thus d₋ + d₊ = 2c|X|/a. Combining this sum with the signed difference and substituting either squared distance yields X²/a² − Y²/(c² − a²) = 1. Define b² = c² − a² > 0, giving the horizontal form X²/a² − Y²/b² = 1.

Conversely, that equation implies |X| ≥ a. Substitute Y² = b²(X²/a² − 1) into the distance squares to obtain d₋ = |cX/a + a| and d₊ = |cX/a − a|. Here |cX/a| ≥ c > a. If X is positive their difference is 2a; if X is negative it is −2a. The absolute difference is therefore 2a on both branches. These sign checks recover the original locus after squaring.

The vertices are (h ± a, k), the foci (h ± c, k), and c² = a² + b². The semiaxis a names the transverse, or opening, direction; unlike the ellipse, a need not exceed b. Exchanging the axes gives the vertical form Y²/a² − X²/b² = 1, with vertices (h, k ± a) and foci (h, k ± c). The positive squared term identifies the opening direction after the right side has been normalised to 1.

Hyperbola asymptotes guide a graph but are not part of it

For the horizontal form X²/a² − Y²/b² = 1, the asymptotes are Y = ±(b/a)X, or y − k = ±(b/a)(x − h). For the vertical form Y²/a² − X²/b² = 1, they are Y = ±(a/b)X. These lines pass through the centre. The auxiliary rectangle with half-width a and half-height b in the horizontal case supplies their slopes, but its corners and minor-direction endpoints are not on the hyperbola.

The horizontal equation gives Y = ±(b/a)√(X² − a²) for |X| ≥ a. As |X| increases without bound, the nonnegative gap |X| − √(X² − a²) equals a²/(|X| + √(X² − a²)) and tends to zero. For X tending to positive infinity, the square-root signs approach the two lines Y = ±(b/a)X with matching signs; for X tending to negative infinity, the signs are exchanged. This algebraic argument establishes both asymptotes; a few distant points do not prove the limiting behaviour.

Substituting either asymptote into the standard equation makes the left side zero, so no point on those lines satisfies the right side 1. This hyperbola never crosses its own asymptotes. The assertion follows from this particular equation, rather than a universal claim that every graph must avoid every asymptote.

The full horizontal locus has x ≤ h − a or x ≥ h + a and every real y occurs; the centre and the gap between the vertices are excluded. The vertical locus has y ≤ k − a or y ≥ k + a and every real x occurs. A finite viewing window clips these unbounded branches; the intersections with the window boundary are not endpoints of the mathematical curve.

A parabola uses equal focus and perpendicular line distances

A parabola is the locus of points equally distant from a focus and a directrix line, with the focus not on the line. For a vertical axis, use vertex (h, k), focus (h, k + p) and directrix y = k − p, with p ≠ 0. The focus offset p is signed: its distance from the vertex is |p|. The vertex is halfway between the focus and its perpendicular foot on the directrix.

The defining equality is √(X² + (Y − p)²) = |Y + p|. Both sides are nonnegative, so squaring is reversible and gives X² + (Y − p)² = (Y + p)², hence X² = 4pY. Conversely, this polynomial equality makes the two squared distances equal; their nonnegative square roots recover the original equality. The absolute line distance handles both signs of p correctly.

If p > 0, the vertical parabola opens upward with y ≥ k; if p < 0 it opens downward with y ≤ k. Its axis is x = h. For a horizontal axis, the analogous equation is Y² = 4pX, with focus (h + p, k), directrix x = h − p and axis y = k. It opens right for p > 0 and left for p < 0. The number 4p in the equation is not the focus offset p.

A parabola has a vertex, not a centre of reflection symmetry. In a vertical parabola, setting Y = p finds the focus-level chord endpoints (h ± 2|p|, k + p); its length is 4|p|. For a horizontal parabola the corresponding endpoints are (h + p, k ± 2|p|). These endpoints lie on the curve, whereas the focus itself does not: substituting it into the defining distances gives zero to the focus but a positive distance 2|p| to the directrix.

At p = 0 the focus would lie on the directrix, violating the ordinary-parabola hypothesis. The remaining distance equality has a line as its locus: X² = 0 in the vertical orientation or Y² = 0 in the horizontal orientation. Calling that line an ordinary parabola would hide the missing focus/directrix separation.

Eccentricity compares distances under a nonzero line-distance guard

For a focus F not on a directrix line L and e > 0, consider PF = e·dist(P, L). On this locus the line distance cannot be zero: otherwise PF would also be zero, placing F on L. Thus the equivalent ratio PF/dist(P, L) = e is defined. The distance to L is perpendicular and nonnegative, so a signed coordinate difference must not replace it.

For the horizontal noncircular ellipse or hyperbola, e = c/a. Pair the right focus (h + c, k) with the directrix x = h + a²/c. The locus calculations above give PF = |a − cX/a| = (c/a)|X − a²/c|, so the ratio is e. Pair the left focus with x = h − a²/c in the same way. For vertical forms exchange the coordinate roles; their directrices are y = k ± a²/c. Ellipses have 0 < e < 1 and hyperbolas e > 1, while a parabola's original equal-distance definition gives e = 1.

The ratio also proves the classification in a suitable axis-aligned frame. Put F = (f, 0), L: u = d, with f ≠ d. Squaring √((u − f)² + v²) = e|u − d| is reversible. For e ≠ 1, completing the square gives (1 − e²)(u − H)² + v² = e²(f − d)²/(1 − e²), where H = (f − e²d)/(1 − e²). If 0 < e < 1, both quadratic coefficients and the right side are positive, giving an ellipse after division. If e > 1, multiply by −1 to obtain a positive u-square minus a positive v-square with a positive right side, giving a hyperbola. At e = 1 the equation instead becomes v² = 2(f − d)(u − (f + d)/2), an ordinary parabola because f − d ≠ 0.

A circle has coincident foci c = 0 in the equal-axis ellipse case, so its conventional eccentricity is e = c/a = 0 for a > 0. The directrix formula a²/c is then undefined. A circle of positive radius is not obtained by setting e = 0 with a fixed finite directrix in PF = e·dist(P, L); that equation gives only the focus point. The circular eccentricity is the limiting convention as the two foci coincide, rather than a finite directrix construction.

Classify boundary data before using a standard form

Distance inequalities explain the exclusions. For two foci separated by 2c > 0, the distance sum is at least 2c, with equality exactly on their connecting segment. Thus sum 2a < 2c gives an empty set, equality gives the segment, and 2a > 2c gives the ordinary ellipse. For the absolute distance difference, the reverse triangle inequality gives |d₋ − d₊| ≤ 2c.

A difference constant strictly between 0 and 2c gives an ordinary hyperbola. A zero difference gives the perpendicular bisector of the focus segment. A difference of exactly 2c gives the two exterior rays on the focus line, including their focus endpoints; a larger constant gives an empty set. The equality cases follow from equality in the triangle inequality, which requires the relevant points to be collinear in the appropriate order.

If the foci coincide, their distance sum is twice the distance to their common point. A positive sum gives a circle and zero sum gives just that point; a negative sum is impossible. Their absolute distance difference is zero at every point in the plane. Thus a zero difference constant gives the entire plane, while any nonzero constant gives an empty set. These cases are excluded by the distinct-focus hypotheses of an ordinary noncircular ellipse or hyperbola.

A normalised polynomial equation and an original parameter description need separate checks. For example, X² + Y² = R gives a circle, point or empty set according to R's sign. In contrast, a claimed positive-radius construction with nonpositive radius fails its original hypothesis. Likewise, X²/a² + Y²/b² = 1 with a = 0 or b = 0 contains division by zero. Clearing the denominators does not make that original expression defined: a limiting equation may even include points outside the original distance-defined boundary case.

Complete squares and normalise the nonzero right side before reading signs and denominators. For example, −X²/9 + Y²/4 = −1 becomes X²/9 − Y²/4 = 1, a horizontal hyperbola. With right side zero, X²/9 − Y²/4 = 0 instead factors into the two lines Y = ±(2/3)X. A zero or negative normalising constant can therefore change the classification; it is not merely a cosmetic scaling choice.

Exact parameterisations explain what a drawing leaves out

A circle can be parameterised by x = h + r cos t, y = k + r sin t for every real t. A horizontal ellipse similarly uses x = h + a cos t, y = k + b sin t; exchange a and b for a vertical ellipse. Substitution verifies every such point. Conversely, dividing a point's translated coordinates by the positive semiaxes gives a unit-circle point, which has some real angle t. This proves that the parameterisation reaches the full locus, rather than only selected special angles.

For the horizontal hyperbola, a complete two-branch parameterisation is X = σa(1 + t²)/(1 − t²), Y = 2bt/(1 − t²), with σ either 1 or −1 and −1 < t < 1. The denominator is positive on this open interval. Substitution gives the standard equation. Conversely, put w = |X|/a ≥ 1 and z = Y/b; then w² − z² = 1. The inverse t = z/(w + 1) lies in (−1, 1) and returns those coordinates with σ = sign X. Both signs are needed for both branches; t = ±1 is excluded and does not represent a finite endpoint.

For a vertical parabola, x = h + t, y = k + t²/(4p), with p ≠ 0 and every real t, reaches every point because t is exactly X. Exchange the coordinate roles for the horizontal form. This also gives the exact coordinate range in the opening direction and its unbounded extent. A parameter restricted to a small interval displays only part of the curve.

Read centres, foci, directrices and asymptotes as labelled reference objects, not as boundary points unless an exact substitution proves otherwise. A screen can distort a circle into an apparent ellipse when its axes have different scales. Equal units and exact feature tables let you interpret the geometry consistently. Connecting a few valid points with straight segments creates a different set; a smooth sampled curve is a visual approximation, and clipped hyperbola or parabola pieces continue beyond the window.

Compare exact loci, their features and guarded point checks

Start with the translated five-unit circle. Predict whether the centre belongs, then compare its zero distance with the boundary point’s distance 5. Change to the completed-square point and empty cases: what does a radius square of zero or a negative value allow? Finally choose the negative original distance target and explain why a point can satisfy its squared consequence while failing the original condition.

In the horizontal five–four ellipse, compare the two distances from P₁ to the foci. Their sum is fixed even though the distances differ. Predict what happens when the required sum shrinks to the focus separation, then switch to that boundary case and test an interior point of the resulting segment. A still smaller sum has no solution. Coincident foci with a positive sum give a circle; coincident foci with zero sum give one point.

For the horizontal four–three hyperbola, compare P₁ on the right branch with P₂ on the left. Which signed distance difference changes, and which absolute difference stays the same? Then select a case where b > a to test whether the transverse semiaxis must be the larger one. The maximal, excessive and zero-difference cases let you compare two exterior rays, an empty locus, the perpendicular bisector of distinct foci and the entire plane for coincident foci.

For the left-opening parabola, follow the perpendicular segment from P₁ to the directrix rather than a sloping line to another point on it. Predict the opening direction after changing the sign of p. In a zero-offset boundary case, inspect the focus/directrix intersection: both distances are zero, so the equality holds, but their ratio is undefined. The separate focus (2, 0) and directrix x = 8 case has vertex (5, 0) and p = −3.

The panel uses u for the translated axial coordinate and v for the perpendicular coordinate: u = x − h, v = y − k for a horizontal axis, and u = y − k, v = x − h for a vertical axis. Exact tables retain reference features, nonnegative distances, guarded ratios and point membership, alongside the complete x-domain, y-range and parameter scope. The equal-unit frame shows analytic bounded circles and ellipses, closed segments and point loci, or clipped approximate traces with continuation arrows for unbounded curves, lines and rays. Empty and entire-plane cases have explicit symbolic classifications; their visible reference objects do not become locus boundaries.

An exact point check answers whether the selected point belongs to the stated locus. Complete distance identities and the parameter arguments establish the full curve. Finite displayed cases and plotting windows illustrate those proofs; they do not replace the general implications, denominator guards or continuation of unbounded branches.

Worked examples

Try it yourself

1. Complete squares in x² + y² + 8x − 6y = 0. Find the centre and radius, and test whether P = (−1, 7) and the centre itself are on the circle.

   Show the answer and reasoning

   The equation is (x + 4)² + (y − 3)² = 25, with centre (−4, 3) and radius 5. P is on the circle; the centre is inside its disk and is not on the circle.

   Add 16 and 9 to both sides while completing the squares. At P the translated coordinates are 3 and 4, giving 3² + 4² = 25. At the centre they are both zero, giving 0 rather than 25. The positive right side gives an ordinary circle, and the radius is its positive square root.

2. For (x + 2)²/9 + (y − 1)²/25 = 1, identify the centre, major direction, vertices, minor-axis endpoints, foci, eccentricity and directrices.

   Show the answer and reasoning

   Centre (−2, 1); vertical major axis; vertices (−2, −4), (−2, 6); minor endpoints (−5, 1), (1, 1); foci (−2, −3), (−2, 5); e = 4/5; directrices y = −21/4 and y = 29/4.

   The larger denominator 25 is under the Y-square, so a = 5 and b = 3. Then c² = 25 − 9 = 16 and c = 4. Translate the positive and negative vertical offsets from the centre for vertices and foci, and the horizontal minor offsets for the other endpoints. The paired directrices are y = k ± a²/c = 1 ± 25/4. The denominator values are squared semiaxes.

3. For x²/4 − y²/25 = 1, find its vertices, foci, eccentricity and asymptotes. Is it invalid because the negative-term denominator is larger, and are (0, ±5) on the curve?

   Show the answer and reasoning

   Vertices (±2, 0); foci (±√29, 0); e = √29/2; asymptotes y = ±(5/2)x. The hyperbola is valid with a = 2 and b = 5. The reference points (0, ±5) are not on it.

   A hyperbola requires a > 0 and b > 0, not a > b. Its positive X-square fixes the opening direction and a = 2. Use c² = a² + b² = 29, then e = c/a > 1 and slopes ±b/a. At (0, ±5), the left side is −1 rather than 1; those auxiliary-rectangle reference points are not boundary points.

4. Find the equation for focus (2, −1) and directrix y = 3. Give the vertex, p and opening direction, then check P = (6, −1) using the original distances.

   Show the answer and reasoning

   Vertex (2, 1), p = −2, equation (x − 2)² = −8(y − 1), opening downward. P belongs because its distances to the focus and the directrix are both 4.

   The perpendicular focus foot on the directrix is (2, 3), whose midpoint with the focus is (2, 1). Hence p = −1 − 1 = −2 and 4p = −8. At P, the focus distance is 4 horizontally and the directrix distance is |−1 − 3| = 4. Substitution also gives 16 = −8(−2), and the curve satisfies y ≤ 1.

5. For focus F = (2, 0), directrix x = 8 and eccentricity e = 1/2, derive the locus equation and its centre and semiaxes. Check the ratio at P = (4, 0).

   Show the answer and reasoning

   The locus is x²/16 + y²/12 = 1, with centre (0, 0), major semiaxis 4 and minor semiaxis 2√3. At P, PF = 2 and dist(P, L) = 4, so the ratio is 1/2.

   The original equality is √((x − 2)² + y²) = |x − 8|/2. Squaring and expanding gives (3/4)x² + y² = 12; the linear terms cancel. Divide by 12 to obtain the equation. The focus is not on the line and both sides of the equality are nonnegative, so the squared equation is equivalent. The semiaxes are positive square roots of 16 and 12, and c² = 16 − 12 = 4 agrees with the given focus offset 2.

6. Two distinct foci are (−3, 0) and (3, 0). Classify distance sums 6 and 4, and absolute distance differences 0, 6 and 8. Also classify x² + y² = 0, x² + y² = −1 and y² = 0 without calling them ordinary conics.

   Show the answer and reasoning

   Sum 6: the closed segment from (−3, 0) to (3, 0); sum 4: empty. Difference 0: line x = 0; difference 6: rays y = 0 with x ≤ −3 or x ≥ 3; difference 8: empty. The final equations describe respectively the point (0, 0), an empty real set and the line y = 0.

   The focus separation is 6. Triangle inequality forces the sum to be at least 6, with equality precisely between the foci. Reverse triangle inequality bounds the absolute difference by 6, with equality on the two exterior rays; equal focus distances instead give the perpendicular bisector. A sum of two squares is zero only when both are zero and cannot be negative. The single equation y² = 0 is equivalent to y = 0 and leaves x unrestricted.

7. For x²/9 + y²/4 = 1, give a complete trigonometric parameterisation and an additional exact point besides the four axis endpoints. Does joining those four endpoints by straight segments reproduce the ellipse? Why are a hyperbola's viewing-window intersections not endpoints?

   Show the answer and reasoning

   Use x = 3cos t, y = 2sin t for all real t. One further point is (9/5, 8/5). Straight joining segments do not reproduce the ellipse. A finite window clips a hyperbola whose exact branches continue without bound.

   Substitution and cos² t + sin² t = 1 put every parameter value on the ellipse; conversely, every ellipse point normalises to a unit-circle point and hence has some angle t. At (9/5, 8/5), the equation gives 9/25 + 16/25 = 1. The midpoint (3/2, 1) of the straight chord between (3, 0) and (0, 2) gives 1/4 + 1/4 = 1/2, so that chord is not the locus. The hyperbola's exact coordinate equation and its parameterisation with unbounded coordinate values, rather than a screen edge, determine its extent.

What to carry forward

• A conic locus includes every point satisfying its definition. Exact point checks illustrate membership; complete two-direction implications and parameter arguments establish the whole set.
• A circle has positive radius and equation (x − h)² + (y − k)² = r². Completing squares requires a positive resulting radius square for an ordinary circle; zero gives a point and a negative value an empty real set.
• A noncircular ellipse has distance sum 2a greater than focus separation 2c, with a > b > 0 and c² = a² − b². Its larger denominator marks the major-axis direction, and full axis lengths are twice the semiaxes.
• A hyperbola uses the absolute distance difference 2a, with 0 < a < c and c² = a² + b². Both branches are required; its positive normalised square marks the opening direction, and a need not exceed b.
• A parabola has a focus off its directrix. A signed nonzero p determines its orientation and focus offset, while distances use |p| and absolute perpendicular line differences; its vertex is not a centre.
• Eccentricity is a positive focus-to-directrix distance ratio for the noncircular finite-directrix construction: ellipse e < 1, parabola e = 1, hyperbola e > 1. A circle's conventional e = 0 is the coincident-focus limit, with no finite directrix from a²/c.
• Retain original hypotheses and nonzero denominators when classifying boundary data. Degenerate segments, lines, rays, points and empty sets differ from ordinary curves, while sampled or clipped drawings differ from the full exact loci.