Conic Sections | Ronny Best Mathematics — CBC Senior Secondary Kenya

Section 1 · Introduction

What Are Conic Sections?

Foundational

A conic section is any curve obtained by cutting a right circular cone with a flat plane. The angle and position of the cut determines which of the four curves results: a circle, ellipse, parabola, or hyperbola.

Origin and History

The Greek mathematician Apollonius of Perga (c. 262–190 BC) wrote an eight-volume treatise titled Conics, in which he studied all four types of conic sections systematically — a remarkable achievement for the ancient world. The names "parabola," "ellipse," and "hyperbola" are all due to Apollonius, derived from Greek words meaning "application," "deficiency," and "excess" respectively.

Conics are not merely abstract curves. Johannes Kepler (1609) discovered that planets follow elliptical orbits around the Sun. Galileo Galilei showed that projectiles follow parabolic paths. Satellite dishes, car headlight reflectors, and suspension bridges all employ the geometric properties of parabolae. Hyperbolic navigation systems (LORAN) were used to guide ships before GPS.

Visualising the Cuts

Imagine a double cone — two identical cones placed tip to tip. A flat plane cuts through this double cone in different ways:

⭕ Circle

Plane cuts perpendicular to the cone's axis. Every point on the curve is equidistant from the centre. Eccentricity e = 0.

🥚 Ellipse

Plane cuts at an angle, passing through only one nappe (half-cone). A stretched circle. Eccentricity 0 < e < 1.

⌒ Parabola

Plane is parallel to exactly one slant side of the cone. Opens infinitely in one direction. Eccentricity e = 1.

↔ Hyperbola

Plane cuts through both nappes of the double cone. Produces two separate branches. Eccentricity e > 1.

The General Second-Degree Equation

Every conic section can be expressed as a second-degree (quadratic) equation in two variables. The general form is:

General Second-Degree Equation

Ax² + Bxy + Cy² + Dx + Ey + F = 0

The nature of the conic is determined by the discriminant B² − 4AC:

B²−4AC < 0

Circle (if A=C, B=0) or Ellipse

e.g. x² + y² = 25 or x²/9 + y²/4 = 1

B²−4AC = 0

Parabola

e.g. y = x² or x² = 4y

B²−4AC > 0

Hyperbola

e.g. x²/9 − y²/16 = 1

Section 2 · Unifying Concept

The Focus-Directrix Definition & Eccentricity

Senior Secondary

All four conics can be defined by a single, elegant relationship involving a fixed point (the focus), a fixed line (the directrix), and a constant ratio called the eccentricity (e). This unified definition shows that the four conics are not separate curves — they are one family of curves, distinguished only by the value of e.

Focus-Directrix Definition

Definition For any point P on the conic: PF / Pd = e (constant)

PF Distance from P to the focus F

Pd Perpendicular distance from P to the directrix

e = 0 Circle (directrix at infinity)

0 < e < 1 Ellipse — P is always closer to the focus than to the directrix

e = 1 Parabola — P is equidistant from focus and directrix

e > 1 Hyperbola — P is always farther from the directrix than from the focus

Why eccentricity matters: The word "eccentric" literally means "off-centre." Eccentricity measures how much a conic deviates from being a perfect circle. A circle has zero eccentricity (perfectly centred). As e increases toward 1, an ellipse becomes more and more elongated. At e = 1 the curve opens up into a parabola, and for e > 1 the curve splits into two branches — a hyperbola.

Section 3 · First Conic

The Circle

Senior Secondary

A circle is the set of all points in a plane that are equidistant from a fixed point called the centre. That fixed distance is called the radius. It is the special case of an ellipse where both axes are equal.

Standard Form and General Form

Circle Equations

Standard form (x − h)² + (y − k)² = r² — centre (h, k), radius r

General form x² + y² + 2gx + 2fy + c = 0

Centre from general (−g, −f)

Radius from general r = √(g² + f² − c) — must have g² + f² − c > 0

Tangent at (x₁, y₁) x·x₁ + y·y₁ = r² (for circle centred at origin)

Length of tangent from (h, k) √(h² + k² − r²)

Converting General to Standard Form

The key technique is completing the square for both x and y. Group x-terms together and y-terms together, complete the square on each group, and move constants to the right-hand side.

Completing the square reminder: To complete the square for x² + bx: take half the coefficient of x, square it, and add it to both sides. x² + bx + (b/2)² = (x + b/2)²

Worked Examples — Circle

Worked Examples · Section 3

Example 1 Convert x² + y² − 6x + 4y − 3 = 0 to standard form. State the centre and radius.

Group

Group x-terms and y-terms: (x² − 6x) + (y² + 4y) = 3

Complete x

Half of −6 is −3; (−3)² = 9. (x² − 6x + 9) + (y² + 4y) = 3 + 9

Complete y

Half of 4 is 2; (2)² = 4. (x − 3)² + (y² + 4y + 4) = 3 + 9 + 4 = 16

Write

(x − 3)² + (y + 2)² = 16

State

Centre: (3, −2) Radius: r = √16 = 4

Centre (3, −2) · Radius 4

Example 2 Find the equation of the tangent to x² + y² = 50 at the point (5, 5).

Verify

Check (5, 5) is on the circle: 5² + 5² = 25 + 25 = 50 ✓

Tangent formula

For x² + y² = r², tangent at (x₁, y₁) is: x·x₁ + y·y₁ = r²

Substitute

x(5) + y(5) = 50 ⟹ 5x + 5y = 50 ⟹ x + y = 10

Tangent: x + y = 10

Example 3 Find the equation of the circle with centre (−2, 3) passing through (1, 7).

Radius

r = distance from centre to point = √((1−(−2))² + (7−3)²) = √(9 + 16) = √25 = 5

Equation

(x − (−2))² + (y − 3)² = 5²

Simplify

(x + 2)² + (y − 3)² = 25

(x + 2)² + (y − 3)² = 25

Example 4 Find the length of the tangent drawn from point (6, 4) to the circle x² + y² − 2x − 4y − 4 = 0.

Standard form

Complete the square: (x−1)² + (y−2)² = 9. Centre (1, 2), radius 3.

Distance

Distance from (6, 4) to centre (1, 2): d = √((6−1)² + (4−2)²) = √(25+4) = √29

Tangent length

Length = √(d² − r²) = √(29 − 9) = √20 = 2√5

Length of tangent = 2√5

Section 4 · Second Conic

The Parabola

Senior Secondary

A parabola is the set of all points in a plane equidistant from a fixed point (the focus) and a fixed line (the directrix). It is the conic with eccentricity exactly equal to 1. Parabolas appear everywhere: satellite dishes, car headlights, suspension bridges, and projectile motion all follow parabolic paths.

Key Terminology

Every parabola has a vertex (the turning point, midway between focus and directrix), an axis of symmetry (the line through the focus perpendicular to the directrix), and a latus rectum (the chord through the focus parallel to the directrix, with length 4a — this measures how "wide" the parabola is at the focus).

Standard Forms of Parabola (vertex at origin)

Opens upward x² = 4ay — focus (0, a), directrix y = −a, axis along y

Opens downward x² = −4ay — focus (0, −a), directrix y = a

Opens right y² = 4ax — focus (a, 0), directrix x = −a, axis along x

Opens left y² = −4ax — focus (−a, 0), directrix x = a

General vertex (h, k) (x − h)² = 4a(y − k) or (y − k)² = 4a(x − h)

Latus rectum length |4a| (the width of the parabola at the focus)

Reflective property: Any ray parallel to the axis of a parabola is reflected through the focus. This is why parabolic dishes concentrate all incoming signals (radio waves, light, sound) at the focus, and why car headlights place the bulb at the focus to produce a parallel beam.

Worked Examples — Parabola

Worked Examples · Section 4

Example 5 For the parabola y² = 20x, find the focus, directrix, vertex, and latus rectum length.

Identify

Form: y² = 4ax ⟹ 4a = 20 ⟹ a = 5

Elements

Vertex: (0, 0) Focus: (5, 0) Directrix: x = −5

Latus rectum

Length = 4a = 20 (chord through focus, parallel to directrix)

Vertex (0,0) · Focus (5,0) · Directrix x = −5 · Latus rectum 20

Example 6 Find the equation of the parabola with focus (0, −3) and directrix y = 3.

Orientation

Focus is below the directrix, so the parabola opens downward.

Value of a

The vertex is midway between focus and directrix: vertex = (0, 0). a = 3.

Equation

Opens downward: x² = −4ay = −4(3)y = −12y ⟹ x² = −12y

x² = −12y

Example 7 Convert (x − 2)² = 8(y + 1) to standard form and identify the vertex, focus, and directrix.

Identify form

This is (x − h)² = 4a(y − k) with h = 2, k = −1, 4a = 8 ⟹ a = 2

Vertex

Vertex = (h, k) = (2, −1)

Focus

Opens upward (4a positive). Focus = (h, k + a) = (2, 1)

Directrix

y = k − a = −1 − 2 = y = −3

Vertex (2, −1) · Focus (2, 1) · Directrix y = −3

Section 5 · Third Conic

The Ellipse

Senior Secondary

An ellipse is the set of all points in a plane for which the sum of the distances from two fixed points (the foci) is constant. It is a "flattened circle" — eccentricity between 0 (circle) and 1. Planetary orbits, the shape of a rugby ball, and whispering galleries (where sound from one focus travels to the other) are all ellipses.

Key Terms and Relationships

In an ellipse x²/a² + y²/b² = 1 (with a > b > 0, major axis along x), the semi-major axis is a, the semi-minor axis is b, and the distance from the centre to each focus is c, where c² = a² − b². This relationship is fundamental: memorise it. Note carefully that c² = a² − b² (subtract for ellipse), unlike a² = b² + c² for right triangles or c² = a² + b² for hyperbolas.

Standard Ellipse — Major Axis Along x (a > b > 0)

Equation x²/a² + y²/b² = 1

Centre (0, 0)

Vertices (ends of major) (±a, 0)

Co-vertices (ends of minor) (0, ±b)

Foci (±c, 0) where c² = a² − b²

Eccentricity e = c/a (0 < e < 1)

Latus rectum length 2b²/a

Tangent at (x₁, y₁) x·x₁/a² + y·y₁/b² = 1

Standard Ellipse — Major Axis Along y (b > a, or swap a and b)

Equation x²/b² + y²/a² = 1 (a is always the larger denominator)

Vertices (0, ±a) Foci: (0, ±c)

Worked Examples — Ellipse

Worked Examples · Section 5

Example 8 For the ellipse x²/25 + y²/9 = 1, find: a, b, c, the foci, eccentricity, and latus rectum length.

Read off

a² = 25 ⟹ a = 5. b² = 9 ⟹ b = 3. (a > b, so major axis along x.)

Find c

c² = a² − b² = 25 − 9 = 16 ⟹ c = 4

Foci

Foci: (±4, 0) = (4, 0) and (−4, 0)

Eccentricity

e = c/a = 4/5 = 0.8

Latus rectum

2b²/a = 2(9)/5 = 18/5 = 3.6

a=5, b=3, c=4 · Foci (±4, 0) · e=0.8 · LR=3.6

Example 9 Find the equation of the ellipse with foci (±5, 0) and semi-major axis a = 7.

Given

c = 5 (foci on x-axis), a = 7

Find b²

b² = a² − c² = 49 − 25 = 24

Equation

x²/49 + y²/24 = 1

x²/49 + y²/24 = 1

Example 10 Find the tangent to x²/16 + y²/9 = 1 at the point (2, 3√3/2).

Tangent formula

x·x₁/a² + y·y₁/b² = 1 with a² = 16, b² = 9, x₁ = 2, y₁ = 3√3/2

Substitute

x(2)/16 + y(3√3/2)/9 = 1 ⟹ x/8 + y√3/6 = 1

Simplify

Multiply through by 24: 3x + 4√3 y = 24

3x + 4√3 y = 24

Section 6 · Fourth Conic

The Hyperbola

Senior Secondary

A hyperbola is the set of all points in a plane for which the absolute difference of the distances from two fixed points (the foci) is constant. Unlike the other conics, a hyperbola has two separate branches and two asymptotes — lines the branches approach but never cross.

Critical Distinction: Ellipse vs Hyperbola

Ellipse

c² = a² − b² (subtract)

Sum of distances = constant

e = c/a, 0 < e < 1

Hyperbola

c² = a² + b² (add)

Difference of distances = constant

e = c/a, e > 1

Standard Hyperbola — Transverse Axis Along x

Equation x²/a² − y²/b² = 1

Vertices (±a, 0)

Foci (±c, 0) where c² = a² + b²

Eccentricity e = c/a > 1

Asymptotes y = ±(b/a)x — lines the branches approach but never reach

Latus rectum length 2b²/a

Hyperbola — Transverse Axis Along y

Equation y²/a² − x²/b² = 1

Vertices (0, ±a) Foci: (0, ±c)

Asymptotes y = ±(a/b)x (note: a over b, not b over a)

Rectangular Hyperbola (special case: a = b)

Equation xy = c² (asymptotes are the coordinate axes)

Parametric form x = ct, y = c/t (t ≠ 0)

Tangent at (ct, c/t) x + t²y = 2ct

Worked Examples — Hyperbola

Worked Examples · Section 6

Example 11 For x²/9 − y²/16 = 1, find a, b, c, foci, eccentricity, and asymptotes.

Read off

a² = 9 ⟹ a = 3. b² = 16 ⟹ b = 4. Transverse axis along x (x² is positive).

Find c

c² = a² + b² = 9 + 16 = 25 ⟹ c = 5

Foci & e

Foci: (±5, 0) e = c/a = 5/3 ≈ 1.67

Asymptotes

y = ±(b/a)x = ±(4/3)x ⟹ y = (4/3)x and y = −(4/3)x

a=3, b=4, c=5 · Foci (±5, 0) · e=5/3 · Asymptotes y=±(4/3)x

Example 12 Find the equation of the hyperbola with foci (±13, 0) and vertices (±5, 0).

Given

a = 5 (vertices), c = 13 (foci), transverse axis along x.

Find b²

b² = c² − a² = 169 − 25 = 144

Equation

x²/25 − y²/144 = 1

x²/25 − y²/144 = 1

Example 13 Find the tangent to the rectangular hyperbola xy = 12 at the point (2, 6).

Parametric

c² = 12 ⟹ c = 2√3. At (2, 6): ct = 2, so t = 2/(2√3) = 1/√3

Verify

Check: xy = 2 × 6 = 12 ✓ Point is on the hyperbola.

Direct method

For xy = c², differentiate: y + x(dy/dx) = 0 ⟹ dy/dx = −y/x = −6/2 = −3

Tangent line

y − 6 = −3(x − 2) ⟹ y = −3x + 12 ⟹ 3x + y = 12

Tangent: 3x + y = 12

Section 7 · Exam Preparation

Common Errors to Avoid in Examinations

All Levels

These errors appear repeatedly in KCSE scripts and Senior Secondary assessments on conic sections. Study each one carefully — the distinction between ellipse and hyperbola formulas is a particularly common source of lost marks.

✗ Wrong For ellipse: b² = a² + c² (adding)

✓ Correct For ellipse: c² = a² − b² (subtracting, because c < a)

The foci are inside the ellipse, so c < a. Always subtract to find c for an ellipse.

✗ Wrong For hyperbola: c² = a² − b²

✓ Correct For hyperbola: c² = a² + b² (adding, because c > a)

The foci are outside the hyperbola's vertices, so c > a. Add to find c for a hyperbola.

✗ Wrong Saying hyperbola y = ±(a/b)x when transverse axis is along x

✓ Correct x²/a² − y²/b² = 1 has asymptotes y = ±(b/a)x (b over a)

The asymptotes for x²/a² − y²/b² = 1 are y = ±(b/a)x. For y²/a² − x²/b² = 1 they are y = ±(a/b)x.

✗ Wrong Completing the square: adding the constant only to one side

✓ Correct Whatever you add to the left, you must add to the right as well

Both sides of the equation must remain equal. Add (b/2)² to both sides simultaneously.

✗ Wrong Parabola opens "right" because the equation is x² = 4ay

✓ Correct x² = 4ay opens upward (or downward); y² = 4ax opens right (or left)

The variable that is squared determines the axis of symmetry. x² means axis is vertical; y² means axis is horizontal.

✗ Wrong Identifying a as the smaller denominator in x²/9 + y²/25 = 1

✓ Correct a² is always the larger denominator. Here a² = 25, b² = 9, major axis along y.

In an ellipse, a is always the semi-major axis (larger), and the major axis runs along the variable with the larger denominator.

✗ Wrong Length of tangent from (h,k) = h² + k² − r²

✓ Correct Length = √(h² + k² − r²) — must take the square root

The Pythagorean theorem gives the length of the tangent; the formula h² + k² − r² is the squared length.

Section 8 · Quick Reference

Complete Formula Sheet — Conic Sections

Reference

Use this formula sheet during revision. All formulas are required knowledge for the KCSE Senior Secondary Mathematics examination.

Circle

(x−h)² + (y−k)² = r²

General: x²+y²+2gx+2fy+c=0

Centre: (−g, −f)

Radius: √(g²+f²−c)

Circle: Tangent & Related

Tangent at (x₁,y₁): xx₁+yy₁=r²

Tangent length from (h,k): √(h²+k²−r²)

Eccentricity: e = 0

Parabola (vertex origin)

y²=4ax: focus (a,0), dir x=−a

x²=4ay: focus (0,a), dir y=−a

Latus rectum = 4a

Eccentricity: e = 1

Ellipse

x²/a² + y²/b² = 1 (a > b)

Foci: (±c, 0), c² = a²−b²

e = c/a < 1

Latus rectum = 2b²/a

Ellipse: Tangent

xx₁/a² + yy₁/b² = 1

Sum of focal distances = 2a

Directrices: x = ±a/e

Hyperbola

x²/a² − y²/b² = 1

Foci: (±c, 0), c² = a²+b²

e = c/a > 1

Asymptotes: y = ±(b/a)x

Rectangular Hyperbola

xy = c²

Parametric: (ct, c/t)

Tangent: x + t²y = 2ct

Asymptotes: x=0 and y=0

Eccentricity Summary

Circle: e = 0

Ellipse: 0 < e < 1

Parabola: e = 1

Hyperbola: e > 1