7.2 Conic Sections Basics

Key Takeaways

  • A circle with centre (h, k) and radius r has equation (x − h)² + (y − k)² = r²; expanded form x² + y² + Dx + Ey + F = 0 needs equal x² and y² coefficients and no xy term.
  • A parabola is the set of points equidistant from a focus and a directrix; standard forms y² = 4ax and x² = 4ay open along a coordinate axis.
  • An ellipse x²/a² + y²/b² = 1 (a > b) has foci (±ae, 0) where e = √(1 − b²/a²) and major axis length 2a.
  • A hyperbola x²/a² − y²/b² = 1 has foci (±ae, 0) with e = √(1 + b²/a²) and asymptotes y = ±(b/a)x.
  • Useful conic practice includes identifying a conic from an equation, reading centre, radius, or vertices, and applying standard identities.
Last updated: July 2026

Why Conics Matter

Conic sections close out plane analytic geometry in the school-level syllabus. Useful self-checks include identification and one-step formula questions such as “which equation is an ellipse?”, “radius of this circle?”, or “length of the latus rectum?”. Recognise each standard form before attempting longer derivations.

Engineering intuition helps memory: circular arcs approximate wheel and gear sections, parabolic mirrors appear in optics and some antenna dishes, elliptical orbits and stress trajectories show up later in aerospace coursework, and hyperbolic paths model certain fly-by geometries. The exercises use textbook equations rather than mission-design calculations.

Circle

Definition: set of points at fixed distance (radius r) from a fixed centre (h, k).

Standard equation: (x − h)² + (y − k)² = r².

General second-degree form (no xy term, equal coefficients of x² and y²):

x² + y² + Dx + Ey + F = 0,

with centre (−D/2, −E/2) and r = √[(D/2)² + (E/2)² − F], provided the expression under the square root is positive.

FeatureFormula
Centre(h, k) or (−D/2, −E/2)
Radiusr
Diameter endpoints (x₁,y₁), (x₂,y₂)(x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

Worked Example 1 — Completing the square

Find the centre and radius of x² + y² − 6x + 4y − 3 = 0.

Solution: (x² − 6x) + (y² + 4y) = 3 → (x − 3)² − 9 + (y + 2)² − 4 = 3 → (x − 3)² + (y + 2)² = 16.

Centre (3, −2), radius 4.

Worked Example 2 — Equation from centre and point

A circle has centre (−1, 2) and passes through (3, 5). Find its equation.

Solution: r² = (3 − (−1))² + (5 − 2)² = 16 + 9 = 25.

Equation: (x + 1)² + (y − 2)² = 25 → x² + y² + 2x − 4y − 20 = 0.

Parabola

Definition: set of points equidistant from a fixed point (focus) and a fixed line (directrix).

Standard forms with vertex at the origin:

EquationOpensFocusDirectrixLatus rectum
y² = 4ax (a > 0)+x(a, 0)x = −alength 4a
y² = −4ax−x(−a, 0)x = a4a
x² = 4ay+y(0, a)y = −a4a
x² = −4ay−y(0, −a)y = a4a

Eccentricity of every parabola is e = 1.

Worked Example 3 — Focus and directrix

For y² = 12x, find focus, directrix, and latus rectum length.

Solution: 4a = 12 ⇒ a = 3. Focus (3, 0), directrix x = −3, latus rectum 12.

Worked Example 4 — Vertex shifted

Identify the parabola (y − 1)² = −8(x + 2).

Solution: Vertex (−2, 1), 4a = 8 ⇒ a = 2, opens left. Focus: (−2 − 2, 1) = (−4, 1). Directrix: x = −2 + 2 = 0.

Ellipse

Standard form (major axis along x): x²/a² + y²/b² = 1 with a > b > 0.

  • Semi-major axis a, semi-minor axis b.
  • Eccentricity e = √(1 − b²/a²) (0 < e < 1).
  • Foci (±ae, 0); directrices x = ±a/e.
  • Relation: b² = a²(1 − e²).

If the major axis is vertical: x²/b² + y²/a² = 1 (still a > b), foci (0, ±ae).

QuantityValue (major axis on x)
Vertices(±a, 0)
Co-vertices(0, ±b)
Foci(±ae, 0)
Major-axis length2a
Minor-axis length2b
Latus rectum2b²/a

Worked Example 5 — Ellipse parameters

For x²/25 + y²/9 = 1, find e and the foci.

Solution: a² = 25, b² = 9 ⇒ a = 5, b = 3. e = √(1 − 9/25) = √(16/25) = 4/5. Foci (±ae, 0) = (±4, 0).

Worked Example 6 — Which axis is major?

For 9x² + 4y² = 36, rewrite and identify axes.

Solution: Divide by 36: x²/4 + y²/9 = 1. Here a² = 9 along y, b² = 4 along x. Major axis is vertical; a = 3, b = 2; e = √(1 − 4/9) = √(5/9) = √5 / 3; foci (0, ±√5).

Hyperbola

Standard form (transverse axis along x): x²/a² − y²/b² = 1.

  • Eccentricity e = √(1 + b²/a²) (e > 1).
  • Foci (±ae, 0); directrices x = ±a/e.
  • Asymptotes: y = ±(b/a)x.
  • Relation: b² = a²(e² − 1).

Conjugate form y²/a² − x²/b² = 1 opens up/down with asymptotes y = ±(a/b)x.

Featurex²/a² − y²/b² = 1
Vertices(±a, 0)
Foci(±ae, 0)
Asymptotesy = ±(b/a)x
Latus rectum2b²/a

Worked Example 7 — Hyperbola basics

For x²/16 − y²/9 = 1, find e, foci, and asymptotes.

Solution: a = 4, b = 3. e = √(1 + 9/16) = √(25/16) = 5/4. Foci (±5, 0). Asymptotes y = ±(3/4)x.

Recognition Table for Mixed Equations

For Ax² + Bxy + Cy² + Dx + Ey + F = 0 with B = 0 (axis-aligned conics):

Condition (after simplifying)Conic
A = C ≠ 0Circle (if radius real)
A = 0, C ≠ 0 (or A ≠ 0, C = 0)Parabola
A and C same sign, A ≠ CEllipse
A and C opposite signsHyperbola

Worked Example 8 — Identify the conic

Classify 3x² − 2y² + 6x + 4y − 1 = 0.

Solution: Coefficients of x² and y² have opposite signs → hyperbola (complete the square only if the stem asks for centre/vertices).

Common Traps

  1. Ellipse vs circle: equal denominators (or A = C) make a circle; unequal positive denominators make an ellipse.
  2. Hyperbola sign pattern: x²/a² − y²/b² = 1 is not the same as y²/a² − x²/b² = 1—check which variable is positive.
  3. Parabola “4a”: the coefficient of the linear side is 4a, not a; focus distance from vertex is a, not 4a.
  4. Expanded circle: if the x² and y² coefficients differ, it is not a circle in Cartesian orthogonal axes.

Keep the four standard pictures and the eccentricity cut-offs (e = 1 parabola; e < 1 ellipse; e > 1 hyperbola; circle as e = 0 limiting ellipse) ready for rapid MCQ elimination.

Test Your Knowledge

The circle x² + y² + 4x − 6y − 3 = 0 has which centre and radius?

A
B
C
D
Test Your Knowledge

For the parabola y² = −16x, which statement is correct?

A
B
C
D
Test Your Knowledge

For the ellipse x²/169 + y²/144 = 1, what is the eccentricity?

A
B
C
D
Test Your Knowledge

The asymptotes of the hyperbola x²/9 − y²/4 = 1 are:

A
B
C
D