Conic Sections

Circles · Parabolas · Ellipses · Hyperbolas — all four, concept first

MAT 172 — Final Prep
All four conics come from the same source: slice a double cone at different angles. Slice straight across → circle. Slight tilt → ellipse. Parallel to the cone's side → parabola. Through both cones → hyperbola. The equation tells you which slice you're looking at.
ConceptA circle is every point exactly the same distance (radius) from a center point. That's it. The equation is just the Pythagorean theorem: the distance from (x,y) to (h,k) equals r.
(x − h)² + (y − k)² = r²
Center: (h, k) · Radius: r · All points equidistant from center
Interactive circle — drag the sliders
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Equation
x²+y²=9
Circumference
Area
Find the center and radius of x² + y² − 6x + 4y − 3 = 0
1
Group x and y terms: (x² − 6x) + (y² + 4y) = 3
2
Complete the square for x: take half of −6 = −3, square it = 9. Add to both sides: (x² − 6x + 9) + (y² + 4y) = 3 + 9
3
Complete the square for y: take half of 4 = 2, square it = 4. Add to both sides: (x − 3)² + (y + 2)² = 3 + 9 + 4 = 16
4
Standard form: (x − 3)² + (y + 2)² = 16. So r² = 16 → r = 4.
Center: (3, −2) · Radius: 4
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Completing the square — step by step
For x² + bx: take half of b, square it, add to both sides.

Example: x² − 6x → half of −6 = −3 → (−3)² = 9 → write (x − 3)² and add 9 to the right side.

The most common mistake: forgetting to add the same number to both sides. If you add 9 to the left in the parentheses, you must add 9 to the right.
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General form → standard form (the full process)
Ax² + Ay² + Cx + Dy + E = 0 (when A = 1):

1. Move E to the right side
2. Group: (x² + Cx) + (y² + Dy) = −E
3. Complete the square for each group
4. Factor into (x − h)² + (y − k)² = r²

If r² comes out negative → no real circle exists. If r² = 0 → a single point.
ConceptA parabola is every point equidistant from a fixed point (the focus) AND a fixed line (the directrix). The result: a U-shaped curve where the vertex is halfway between focus and directrix. Every satellite dish, headlight reflector, and projectile path is a parabola.
Vertical: y = a(x−h)² + k  ·  Horizontal: x = a(y−k)² + h
Vertex: (h, k) · Opens up/down if vertical, left/right if horizontal · a determines width and direction
Interactive parabola
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Equation
y=x²
Opens
Up
Axis of sym.
x=0
Write y = 2x² − 8x + 5 in vertex form
1
Factor out a from x terms: y = 2(x² − 4x) + 5
2
Complete the square inside: half of −4 = −2, squared = 4. Add and subtract: y = 2(x² − 4x + 4 − 4) + 5
3
Separate: y = 2(x² − 4x + 4) − 2(4) + 5 = 2(x−2)² − 8 + 5
4
Simplify: y = 2(x−2)² − 3
Vertex form: y = 2(x−2)² − 3 · Vertex: (2, −3) · Opens up (a > 0)
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Focus and directrix — what they actually are
For y = a(x−h)² + k, the focus is at (h, k + 1/(4a)) and the directrix is y = k − 1/(4a).

The focus is INSIDE the parabola. The directrix is a line OUTSIDE it on the opposite side. Every point on the parabola is equidistant from both.

Real-world connection: Satellite dishes are parabolas with a receiver at the focus. Parallel signals from space hit the dish and all reflect to the same focus point — that's why the geometry works. Your headlights use the same principle in reverse: a bulb at the focus sends light out in parallel beams.
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Vertical vs. horizontal parabolas
Vertical: y = a(x−h)² + k. The x is squared. Opens up (a > 0) or down (a < 0). Axis of symmetry is vertical: x = h.

Horizontal: x = a(y−k)² + h. The y is squared. Opens right (a > 0) or left (a < 0). Axis of symmetry is horizontal: y = k.

Quick check: which variable is squared? That variable is the one that fans out. The other variable is the axis direction.
ConceptAn ellipse is every point where the SUM of distances to two fixed points (foci) is constant. Stretch a circle and you get an ellipse. The longer axis is called major, the shorter is minor. Planets orbit in ellipses — the sun is at one focus, not the center.
x²/a² + y²/b² = 1  (horizontal)  ·  x²/b² + y²/a² = 1  (vertical)
Center: (h, k) · a = semi-major axis (always larger) · b = semi-minor · c² = a² − b² · Foci at distance c from center
Interactive ellipse
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2.5
c (focus dist)
Orientation
Eccentricity
x²/25 + y²/9 = 1 — find everything
1
Identify: a² = 25 → a = 5. b² = 9 → b = 3. Since 25 > 9, a is under x² → horizontal ellipse.
2
Find c: c² = a² − b² = 25 − 9 = 16 → c = 4.
3
Vertices (endpoints of major axis): (±5, 0). Co-vertices (minor axis endpoints): (0, ±3).
4
Foci: (±4, 0) — along the major axis, c units from center.
Vertices: (±5, 0) · Co-vertices: (0, ±3) · Foci: (±4, 0) · e = 4/5 = 0.8
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How to tell horizontal vs. vertical ellipse
The larger denominator (a²) tells you which axis is major.

• If a² is under x²: major axis is horizontal → ellipse is wider than tall.
• If a² is under y²: major axis is vertical → ellipse is taller than wide.

Foci are always on the major axis — they follow the larger denominator.
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The sum of distances definition (why it works)
For any point P on the ellipse: d(P, F₁) + d(P, F₂) = 2a (always constant).

At the vertex: one distance is a+c and the other is a−c. They add to 2a ✓

Garden ellipse trick: Stick two stakes in the ground (the foci), tie a string of length 2a between them, pull it taut with a pencil, and trace — the pencil draws a perfect ellipse because the string length stays constant.
ConceptA hyperbola is every point where the DIFFERENCE of distances to two foci is constant. Unlike an ellipse (sum), a hyperbola has two separate branches. The asymptotes are the lines the branches approach but never touch — they're your guide for graphing.
Horizontal: x²/a² − y²/b² = 1  ·  Vertical: y²/a² − x²/b² = 1
Center: (h,k) · Vertices: a units from center (along the positive term) · c² = a² + b² · Asymptotes: y = ±(b/a)x
Interactive hyperbola
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0
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c (focus dist)
Asymptotes
Eccentricity
x²/4 − y²/9 = 1 — graph and find everything
1
Identify: x² is positive → horizontal hyperbola. a² = 4 → a = 2. b² = 9 → b = 3.
2
Center: (0, 0). Vertices: (±2, 0) — on the x-axis, a units from center.
3
Draw the central rectangle: width 2a = 4 (horizontal), height 2b = 6 (vertical). Draw asymptotes through corners: y = ±(3/2)x.
4
Find foci: c² = a² + b² = 4 + 9 = 13 → c = √13. Foci: (±√13, 0).
5
Draw two branches: one opening right from vertex (2,0), one opening left from (−2,0), hugging the asymptotes.
Vertices: (±2, 0) · Foci: (±√13, 0) · Asymptotes: y = ±(3/2)x
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The central rectangle method — how to draw asymptotes
1. Draw a rectangle centered at (h, k) with width 2a and height 2b.
2. Draw the diagonals of that rectangle — those ARE the asymptotes.
3. The hyperbola branches open from the vertices and hug the asymptotes.

For horizontal hyperbola: asymptotes are y − k = ±(b/a)(x − h)
For vertical hyperbola: asymptotes are y − k = ±(a/b)(x − h)

Notice: vertical swaps a and b in the asymptote slope. Easy to mess up — always check which term is positive.
Ellipse vs. Hyperbola — the critical differences
FeatureEllipseHyperbola
Sign between terms+ (plus)− (minus)
c² formulac² = a² − b²c² = a² + b²
BranchesOne closed curveTwo open branches
Eccentricity0 < e < 1e > 1
AsymptotesNoneYes — y = ±(b/a)x