Sec 5 SN · Conic Sections · Deep Study
Conic
Sections
Four curves — circle, ellipse, parabola, hyperbola — all come from slicing a cone. This guide connects the geometry to the algebra with diagrams, worked examples, and checkpoints at each step.
📐 6 sections
🔍 19 worked examples
✅ 13 checkpoints
💡 Intuition-first
1
Introduction to Conic Sections
Four curves from one cone
A conic section is what you get when you slice through a double cone at different angles. The angle of the cut determines which curve you see:
| Conic | How to recognise the equation | Key feature |
|---|---|---|
| Circle | (x−h)² + (y−k)² = r² | Equal spread in all directions from centre |
| Ellipse | x²/a² + y²/b² = 1 — two different denominators, both + signs | Two radii (major and minor axes) |
| Parabola | One variable squared: (x−h)² = 4c(y−k) or (y−k)² = 4c(x−h) | Opens in one direction; has focus & directrix |
| Hyperbola | x²/a² − y²/b² = 1 — minus sign between fractions | Two branches; has asymptotes |
⚠️
The sign between the two squared terms is the key distinguisher. Plus sign = circle or ellipse. Minus sign = hyperbola. If only one variable is squared, it's a parabola.
Quick-ID Method
Given a general second-degree equation Ax² + Bxy + Cy² + ... = 0 (with B = 0), check A and C:
A = C (same, non-zero)
Circle
A ≠ C, same sign
Ellipse
A or C = 0
Parabola
A and C opposite signs
Hyperbola
★ Basic
Show solution
Identifying conics from equations
Identify each conic: (a) x² + y² = 25 (b) x²/4 + y²/9 = 1 (c) y² = 12x (d) x²/4 − y²/9 = 1
1
(a) x² + y² = 25
Both variables squared with equal coefficients (+1 each) → Circle with radius 5.
2
(b) x²/4 + y²/9 = 1
Both squared, both positive, different denominators (4 ≠ 9) → Ellipse.
3
(c) y² = 12x
Only one variable (y) is squared → Parabola. Since x is alone (not squared), it opens left or right.
4
(d) x²/4 − y²/9 = 1
Two squared terms with a minus sign between them → Hyperbola. x² is positive → opens left/right.
Answer: (a) Circle (b) Ellipse (c) Parabola (d) Hyperbola
✅ Checkpoint 1
Identify each conic: (a) 4x² + 4y² = 36 (b) 4x² + 9y² = 36 (c) 4x² − 9y² = 36 (d) 4x² = y
(a) 4x² + 4y² = 36 → divide by 4: x² + y² = 9 → Circle, radius 3
(b) 4x² + 9y² = 36 → x²/9 + y²/4 = 1 → Ellipse (different denominators)
(c) 4x² − 9y² = 36 → x²/9 − y²/4 = 1 → Hyperbola (minus sign)
(d) 4x² = y → only x is squared → Parabola, opens upward
2
The Circle
The simplest conic — all points equidistant from the centre
Standard form
(x − h)² + (y − k)² = r²
centre (h, k), radius r > 0
Every point on the circle is exactly r units from the centre (h, k). That's the definition — the equation is just the distance formula squared.
💭
Where does the equation come from?
Distance from (x, y) to (h, k) = √((x−h)² + (y−k)²). Set that equal to r and square both sides: (x−h)² + (y−k)² = r².
Distance from (x, y) to (h, k) = √((x−h)² + (y−k)²). Set that equal to r and square both sides: (x−h)² + (y−k)² = r².
Centre
(h, k)
Radius
r = √(RHS)
Domain
[h−r, h+r]
Range
[k−r, k+r]
Standard ↔ General Form
The general form x² + y² + Dx + Ey + F = 0 hides the centre and radius. To convert, complete the square for both x and y.
★ Basic
Show solution
Writing the equation from centre and radius
Write the equation of a circle with centre (−3, 5) and radius 7.
1
Substitute h = −3, k = 5, r = 7
(x − (−3))² + (y − 5)² = 7²
(x + 3)² + (y − 5)² = 49
Answer: (x + 3)² + (y − 5)² = 49
★★ Medium
Show solution
Converting general to standard form
Find the centre and radius: x² + y² − 4x − 6y + 4 = 0.
1
Group x terms and y terms, move constant
(x² − 4x) + (y² − 6y) = −4
2
Complete the square for x: add (4/2)² = 4 to both sides
(x² − 4x + 4) + (y² − 6y) = −4 + 4 = 0
3
Complete the square for y: add (6/2)² = 9 to both sides
(x² − 4x + 4) + (y² − 6y + 9) = 0 + 9 = 9
4
Factor each group
(x − 2)² + (y − 3)² = 9
Centre: (2, 3). Radius: r = √9 = 3.
Answer: Centre (2, 3), radius 3
★★ Medium
Show solution
Checking whether a point is on, inside, or outside
For (x − 1)² + (y + 2)² = 25, determine the position of: (a) (4, 2) (b) (6, −2) (c) (0, 0)
1
Method: compute (x−1)² + (y+2)² and compare to 25
If = 25 → on circle. If < 25 → inside. If > 25 → outside.
2
(a) Point (4, 2)
(4−1)² + (2+2)² = 9 + 16 = 25 = 25 → ON the circle
3
(b) Point (6, −2)
(6−1)² + (−2+2)² = 25 + 0 = 25 = 25 → ON the circle
4
(c) Point (0, 0)
(0−1)² + (0+2)² = 1 + 4 = 5 < 25 → INSIDE the circle
Answer: (4,2) and (6,−2) are on the circle; (0,0) is inside
★★★ Hard
Show solution
Finding the equation from two known points
A circle has centre (2, −1) and passes through (5, 3). Find the equation.
1
Find the radius — distance from centre to the given point
r = √((5−2)² + (3−(−1))²)
= √(9 + 16)
= √25 = 5
2
Write the equation
(x − 2)² + (y + 1)² = 25
Answer: (x − 2)² + (y + 1)² = 25
✅ Checkpoint 2
Find the centre and radius: x² + y² + 10x − 2y + 1 = 0.
Group: (x² + 10x) + (y² − 2y) = −1
Add (10/2)² = 25 and (2/2)² = 1:
(x + 5)² + (y − 1)² = −1 + 25 + 1 = 25
Centre: (−5, 1), radius: 5
3
The Ellipse
The stretched circle — two foci, constant sum of distances
Horizontal vs Vertical Ellipse
| Feature | Horizontal (a under x²) | Vertical (a under y²) |
|---|---|---|
| Equation form | x²/a² + y²/b² = 1, a > b | x²/b² + y²/a² = 1, a > b |
| Major axis | Along x-axis, length 2a | Along y-axis, length 2a |
| Minor axis | Along y-axis, length 2b | Along x-axis, length 2b |
| Foci | (±c, 0) | (0, ±c) |
| Key relation | c² = a² − b² (a is always the larger value) | |
💡
The foci always live on the major axis (the longer one). The larger denominator tells you which axis is major. If the larger denominator is under x², the major axis is horizontal.
★★ Medium
Show solution
Analysing a centred ellipse
Analyze: x²/25 + y²/9 = 1. Find a, b, c, axes lengths, vertices, and foci.
1
Identify a² and b²
a² = 25 → a = 5 (larger denominator, under x² → horizontal major)
b² = 9 → b = 3
2
Find c
c² = a² − b² = 25 − 9 = 16 → c = 4
3
State all features
Major axis: horizontal, length 2a = 10
Minor axis: vertical, length 2b = 6
Vertices on major axis: (±5, 0)
Co-vertices on minor axis: (0, ±3)
Foci: (±4, 0)
Answer: a=5, b=3, c=4 · Vertices (±5,0) · Foci (±4,0)
★★ Medium
Show solution
Shifted ellipse
Find centre, axes, vertices, and foci of (x−2)²/16 + (y+1)²/7 = 1.
1
Centre and parameters
Centre: (h, k) = (2, −1)
a² = 16 → a = 4 (under x-term → horizontal major)
b² = 7 → b = √7 ≈ 2.65
2
Find c
c² = 16 − 7 = 9 → c = 3
3
Vertices and foci (shift by centre)
Vertices: (2 ± 4, −1) = (−2, −1) and (6, −1)
Co-vertices: (2, −1 ± √7)
Foci: (2 ± 3, −1) = (−1, −1) and (5, −1)
Answer: Centre (2,−1) · Vertices (−2,−1) and (6,−1) · Foci (−1,−1) and (5,−1)
★★★ Hard
Show solution
Writing the equation from geometric data
An ellipse centred at the origin has foci at (0, ±4) and vertices at (0, ±5). Find the equation.
1
Determine orientation
Foci and vertices are on the y-axis → vertical major axis. So a is under y².
2
Read a and c
Vertices at (0, ±5) → a = 5 → a² = 25
Foci at (0, ±4) → c = 4 → c² = 16
3
Find b²
b² = a² − c² = 25 − 16 = 9
4
Write equation (a under y², b under x²)
x²/9 + y²/25 = 1
Answer: x²/9 + y²/25 = 1
✅ Checkpoint 3A
For x²/16 + y²/25 = 1: is the major axis horizontal or vertical? Find the foci.
a² = 25 (larger, under y²) → major axis is vertical.
c² = 25 − 16 = 9 → c = 3
Foci: (0, ±3)
✅ Checkpoint 3B
An ellipse has centre (3, 1), a horizontal major axis with a = 6 and b = 4. Write the equation.
(x − 3)²/36 + (y − 1)²/16 = 1
4
The Conic Parabola
Focus and directrix — a different perspective on the same curve
The conic definition — locus of points equidistant from a focus and a directrix — gives the same curve as y = a(x−h)² + k, just described differently. The key new element is the parameter c: the distance from the vertex to both the focus and the directrix.
📐
Link to quadratic form: In y = a(x−h)² + k, the relationship is a = 1/(4c), so c = 1/(4a). The same parabola, two descriptions.
| Orientation | Equation | Focus | Directrix |
|---|---|---|---|
| Opens UP | (x−h)² = 4c(y−k), c>0 | (h, k+c) | y = k−c |
| Opens DOWN | (x−h)² = −4c(y−k), c>0 | (h, k−c) | y = k+c |
| Opens RIGHT | (y−k)² = 4c(x−h), c>0 | (h+c, k) | x = h−c |
| Opens LEFT | (y−k)² = −4c(x−h), c>0 | (h−c, k) | x = h+c |
★★ Medium
Show solution
Finding focus and directrix
Find the focus and directrix of (x − 2)² = 8(y + 1).
1
Identify form and read parameters
4c = 8 → c = 2
Vertex: (h, k) = (2, −1), opens UP (positive 4c)
2
Focus
(h, k + c) = (2, −1 + 2) = (2, 1)
3
Directrix
y = k − c = −1 − 2 = −3
Answer: Focus (2, 1), directrix y = −3
★★ Medium
Show solution
Horizontal (sideways) parabola
Find vertex, focus, and directrix of (y + 3)² = −12(x − 1).
1
Identify form
y is squared → parabola opens left or right. Negative sign → opens LEFT.
Vertex: (h, k) = (1, −3)
−4c = −12 → c = 3
2
Focus (h − c, k) for leftward
Focus: (1 − 3, −3) = (−2, −3)
3
Directrix (x = h + c) for leftward
Directrix: x = 1 + 3 = 4
Answer: Vertex (1,−3), focus (−2,−3), directrix x = 4
★★★ Hard
Show solution
Converting from standard quadratic form to conic form
Write y = 2x² − 8x + 10 in conic form and identify focus and directrix.
1
Complete the square
y = 2(x² − 4x) + 10
y = 2(x² − 4x + 4 − 4) + 10
y = 2(x − 2)² − 8 + 10
y = 2(x − 2)² + 2
2
Rearrange to conic form (x−h)² = 4c(y−k)
y − 2 = 2(x − 2)²
(x − 2)² = (1/2)(y − 2)
3
Read c
4c = 1/2 → c = 1/8
Vertex: (2, 2), opens UP
4
Focus and directrix
Focus: (2, 2 + 1/8) = (2, 17/8)
Directrix: y = 2 − 1/8 = 15/8
Answer: (x−2)² = (1/2)(y−2) · Focus (2, 17/8) · Directrix y = 15/8
✅ Checkpoint 4
Find the vertex, focus, and directrix of (x + 1)² = −6(y − 4).
Vertex: (−1, 4) [h = −1, k = 4]
Negative sign → opens DOWN
−4c = −6 → c = 3/2
Focus: (−1, 4 − 3/2) = (−1, 5/2)
Directrix: y = 4 + 3/2 = 11/2
5
The Hyperbola
Two branches, asymptotes, and c² = a² + b²
Horizontal vs Vertical Hyperbola
| Feature | Horizontal (x² positive) | Vertical (y² positive) |
|---|---|---|
| Equation | x²/a² − y²/b² = 1 | y²/a² − x²/b² = 1 |
| Opens | Left and right | Up and down |
| Vertices | (±a, 0) | (0, ±a) |
| Foci | (±c, 0) | (0, ±c) |
| Asymptotes | y = ±(b/a)x | y = ±(a/b)x |
| Key relation | c² = a² + b² (c is always larger than a and b) | |
⚠️
Ellipse vs Hyperbola — the critical difference:
Ellipse: c² = a² − b² → c < a < "radius"
Hyperbola: c² = a² + b² → c > a, foci are outside the branches
Ellipse: c² = a² − b² → c < a < "radius"
Hyperbola: c² = a² + b² → c > a, foci are outside the branches
Graphing Strategy — 5 Steps
- Step 1: Identify a, b, c, and orientation (which term is positive).
- Step 2: Draw the central box: width = 2a (horizontal) or 2b (vertical), centred at (h, k).
- Step 3: Draw asymptotes through the box corners: y − k = ±(b/a)(x − h).
- Step 4: Plot vertices on the axis of the positive term.
- Step 5: Sketch branches starting at vertices, curving toward but never touching asymptotes.
★★ Medium
Show solution
Full analysis of a horizontal hyperbola
Analyze: x²/9 − y²/16 = 1.
1
Identify a² and b²
a² = 9 → a = 3 (x² is positive → horizontal, opens left/right)
b² = 16 → b = 4
2
Find c
c² = a² + b² = 9 + 16 = 25 → c = 5
3
Asymptotes
y = ±(b/a)x = ±(4/3)x
4
Summary
Vertices: (±3, 0)
Foci: (±5, 0)
Asymptotes: y = ±(4/3)x
Opens: left and right
Answer: Vertices (±3,0) · Foci (±5,0) · Asymptotes y = ±(4/3)x
★★ Medium
Show solution
Vertical hyperbola
Analyze: y²/4 − x²/12 = 1.
1
Orientation
y² term is positive → vertical hyperbola, opens up and down.
a² = 4 → a = 2 (under y²)
b² = 12 → b = 2√3
2
Find c
c² = 4 + 12 = 16 → c = 4
3
Features
Vertices: (0, ±2)
Foci: (0, ±4)
Asymptotes: y = ±(a/b)x = ±(2/(2√3))x = ±(1/√3)x = ±(√3/3)x
Answer: Opens up/down · Vertices (0,±2) · Foci (0,±4) · Asymptotes y = ±(√3/3)x
★★★ Hard
Show solution
Shifted hyperbola
Find vertices, foci, and asymptotes of (x−1)²/4 − (y+2)²/9 = 1.
1
Centre and parameters
Centre: (h, k) = (1, −2)
a² = 4 → a = 2 (x-term positive → horizontal)
b² = 9 → b = 3
2
Find c
c² = 4 + 9 = 13 → c = √13 ≈ 3.61
3
Shift all features by (1, −2)
Vertices: (1 ± 2, −2) = (−1, −2) and (3, −2)
Foci: (1 ± √13, −2)
Asymptotes: y + 2 = ±(3/2)(x − 1)
Answer: Centre (1,−2) · Vertices (−1,−2) and (3,−2) · Asymptotes y+2 = ±(3/2)(x−1)
✅ Checkpoint 5
For y²/25 − x²/144 = 1: (a) Does it open left/right or up/down? (b) Find vertices. (c) Find c and foci. (d) Write asymptote equations.
(a) y² is positive → opens UP and DOWN
(b) a² = 25 → a = 5 · Vertices: (0, ±5)
(c) c² = 25 + 144 = 169 → c = 13 · Foci: (0, ±13)
(d) Asymptotes: y = ±(a/b)x = ±(5/12)x
6
Completing the Square
Converting from general to standard form
Conics are often given in general form Ax² + Cy² + Dx + Ey + F = 0. Completing the square reveals the type, centre, and parameters. The process is the same for all four conics.
The Algorithm
- Step 1: If A ≠ 1 (or C ≠ 1), factor the coefficient out of its group: e.g. 4(x² − 3x) + ...
- Step 2: Group x-terms together and y-terms together. Move the constant to the right side.
- Step 3: For each group: add (half the linear coefficient)² inside the parentheses, and add the same amount (×any factored-out coefficient) to the right side.
- Step 4: Factor each group as a perfect square binomial.
- Step 5: Divide both sides to get the standard form (= 1 for ellipse/hyperbola).
⚠️
When you factor out a coefficient before completing the square — say 9(x² − 2x + 1) — you must add 9×1 = 9 to the right side, not just 1. The coefficient multiplies what's inside.
★★★ Hard
Show solution
Identifying and converting a hyperbola
Identify the conic and find its key features: 9x² − 4y² − 18x + 16y − 43 = 0.
1
Coefficients: A=9, C=−4 — opposite signs → hyperbola
Group and factor:
9(x² − 2x) − 4(y² − 4y) = 43
2
Complete the square for each group
9(x² − 2x + 1) − 4(y² − 4y + 4) = 43 + 9(1) − 4(4)
9(x − 1)² − 4(y − 2)² = 43 + 9 − 16 = 36
3
Divide by 36
(x − 1)²/4 − (y − 2)²/9 = 1
x² term is positive → horizontal hyperbola, centre (1, 2)
4
Read features
a² = 4 → a = 2; b² = 9 → b = 3
c = √(4 + 9) = √13
Vertices: (1 ± 2, 2) = (−1, 2) and (3, 2)
Foci: (1 ± √13, 2)
Answer: Horizontal hyperbola, centre (1,2), vertices (−1,2) and (3,2)
★★★ Hard
Show solution
Converting a general ellipse
Identify and find features of 4x² + 9y² − 24x + 36y + 36 = 0.
1
A=4, C=9 — same sign, different → ellipse
4(x² − 6x) + 9(y² + 4y) = −36
2
Complete the square
4(x² − 6x + 9) + 9(y² + 4y + 4) = −36 + 4(9) + 9(4)
4(x − 3)² + 9(y + 2)² = −36 + 36 + 36 = 36
3
Divide by 36
(x − 3)²/9 + (y + 2)²/4 = 1
4
Read features
Centre: (3, −2)
a² = 9 → a = 3 (under x → horizontal major)
b² = 4 → b = 2
c² = 9 − 4 = 5 → c = √5
Vertices: (3 ± 3, −2) = (0, −2) and (6, −2)
Foci: (3 ± √5, −2)
Answer: Ellipse, centre (3,−2), vertices (0,−2) and (6,−2), foci (3±√5, −2)
★★ Medium
Show solution
What if the RHS comes out zero or negative?
Analyze: x² + y² − 6x + 8y + 25 = 0.
1
A = C = 1, same sign → circle
(x² − 6x) + (y² + 8y) = −25
2
Complete the square
(x − 3)² + (y + 4)² = −25 + 9 + 16 = 0
3
Interpret the result
r² = 0 means r = 0 — this is a degenerate conic: a single point at (3, −4).
If RHS had been negative, there would be no real solution — an imaginary circle. If RHS = 0, it's a point circle. These are called degenerate conics.
Answer: Degenerate circle — the single point (3, −4)
✅ Checkpoint 6A
Identify the conic and find centre and radius: x² + y² − 6x + 4y − 3 = 0.
Group: (x² − 6x) + (y² + 4y) = 3
Complete: (x − 3)² + (y + 2)² = 3 + 9 + 4 = 16
Circle, centre (3, −2), radius 4
✅ Checkpoint 6B
Complete the square to find the standard form and identify the conic: 16x² + 25y² − 32x + 50y − 359 = 0.
16(x² − 2x) + 25(y² + 2y) = 359
16(x − 1)² + 25(y + 1)² = 359 + 16 + 25 = 400
(x − 1)²/25 + (y + 1)²/16 = 1
Ellipse, centre (1, −1), a=5 (horizontal), b=4