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🎯⭐ INTERACTIVE LESSON

Conic Sections

Learn step-by-step with interactive practice!

Conic Sections - Complete Interactive Lesson

Part 1: Circles

⭕ Introduction to Conic Sections

Part 1 of 7

What Are Conic Sections?

Conic sections are curves formed by intersecting a double cone with a plane:

ConicPlane AngleEquation Type
CirclePerpendicular to axisx2+y2=r2x^2 + y^2 = r^2
EllipseTilted, doesn't hit basex2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
ParabolaParallel to slanty=ax2+bx+cy = ax^2 + bx + c
HyperbolaSteeper than slantx2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

General Second-Degree Equation

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

The discriminant B2−4ACB^2 - 4AC determines the type:

  • B2−4AC<0B^2 - 4AC < 0: ellipse (or circle if A=CA = C, B=0B = 0)
  • B2−4AC=0B^2 - 4AC = 0: parabola
  • B2−4AC>0B^2 - 4AC > 0: hyperbola

⭕ The Circle

Standard Form

(x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2

Center (h,k)(h, k), radius rr.

General Form

x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0

Convert by completing the square.

Example: x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0

Group: (x2−6x)+(y2+4y)=12(x^2 - 6x) + (y^2 + 4y) = 12

Complete: (x2−6x+9)+(y2+4y+4)=12+9+4(x^2-6x+9)+(y^2+4y+4) = 12+9+4

(x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25

Center (3,−2)(3, -2), radius 55.

📐 Circle Properties

Key Facts

  • All points are equidistant from the center
  • A tangent line at point PP is perpendicular to the radius at PP
  • The equation of the tangent at (x0,y0)(x_0, y_0) on circle x2+y2=r2x^2+y^2 = r^2 is: x0x+y0y=r2x_0 x + y_0 y = r^2

Relationship to Other Conics

A circle is a special case of an ellipse where a=ba = b (both semi-axes equal).

Eccentricity of a circle: e=0e = 0.

💡 Degenerate cases: If r2<0r^2 < 0 after completing the square, there is no real circle (empty set). If r2=0r^2 = 0, it is a single point.

Circle Quiz 🎯

Circle Computations 🧮

1) Center of x2+y2−10x+6y+9=0x^2 + y^2 - 10x + 6y + 9 = 0: the xx-coordinate of center = ?

2) Same circle: the yy-coordinate of center = ?

3) Same circle: the radius = ?

Classify Conics 🔽

Exit Quiz ✅

Part 2: Parabolas

🔵 The Ellipse

Part 2 of 7

Standard Form (Center at Origin)

Horizontal major axis: x2a2+y2b2=1(a>b)\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b)

Vertical major axis: x2b2+y2a2=1(a>b)\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \quad (a > b)

Key Elements

ElementHorizontal Major Axis
Vertices(±a,0)(\pm a, 0)
Co-vertices(0,±b)(0, \pm b)
Foci(±c,0)(\pm c, 0) where c2=a2−b2c^2 = a^2 - b^2
Eccentricitye=cae = \frac{c}{a} (with 0<e<10 < e < 1)
Major axis length2a2a
Minor axis length2b2b

Defining Property

The sum of distances from any point on the ellipse to both foci is constant: d1+d2=2ad_1 + d_2 = 2a.

📝 Example: Analyze x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1

a2=25,b2=9  ⟹  a=5,b=3a^2 = 25, b^2 = 9 \implies a = 5, b = 3

c2=25−9=16  ⟹  c=4c^2 = 25 - 9 = 16 \implies c = 4

  • Center: (0,0)(0, 0)
  • Vertices: (±5,0)(\pm 5, 0)
  • Co-vertices: (0,±3)(0, \pm 3)
  • Foci: (±4,0)(\pm 4, 0)
  • Eccentricity: e=45=0.8e = \frac{4}{5} = 0.8

Since ee is close to 1, this ellipse is fairly elongated.

Translated Ellipse

(x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1

Same shape, centered at (h,k)(h, k) instead of the origin.

📊 Eccentricity: Shape of an Ellipse

EccentricityShape
e=0e = 0Circle (foci coincide at center)
e=0.1e = 0.1Nearly circular
e=0.5e = 0.5Moderate oval
e=0.9e = 0.9Very elongated
e→1e \to 1Approaches a line segment

Real-World Eccentricities

  • Earth's orbit: e≈0.017e \approx 0.017 (nearly circular)
  • Mars's orbit: e≈0.093e \approx 0.093
  • Halley's comet: e≈0.967e \approx 0.967 (very elongated)
  • Pluto: e≈0.248e \approx 0.248

💡 A higher eccentricity means the foci are farther from the center relative to the size of the ellipse.

Ellipse Quiz 🎯

Ellipse Calculations 🧮

For x236+y24=1\frac{x^2}{36}+\frac{y^2}{4}=1:

1) aa = ?

2) cc = ? (Enter like "4sqrt2" if needed)

3) Eccentricity e=cae = \frac{c}{a} = ? (Enter as a fraction like "4/6" or simplify)

Ellipse Properties 🔽

Exit Quiz ✅

Part 3: Ellipses

📐 The Parabola

Part 3 of 7

Definition

A parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix).

Standard Forms (Vertex at Origin)

OpensEquationFocusDirectrix
Upx2=4pyx^2 = 4py(0,p)(0, p)y=−py = -p
Downx2=−4pyx^2 = -4py(0,−p)(0, -p)y=py = p
Righty2=4pxy^2 = 4px(p,0)(p, 0)x=−px = -p
Lefty2=−4pxy^2 = -4px(−p,0)(-p, 0)x=px = p

💡 The value pp is the distance from the vertex to the focus (and also from the vertex to the directrix).

📝 Worked Examples

Example 1: Find focus and directrix of x2=12yx^2 = 12y

Compare to x2=4pyx^2 = 4py: 4p=12  ⟹  p=34p = 12 \implies p = 3

  • Opens upward (positive coefficient)
  • Focus: (0,3)(0, 3)
  • Directrix: y=−3y = -3
  • Latus rectum (width through focus): ∣4p∣=12|4p| = 12

Example 2: Write equation with focus at (−2,0)(-2, 0)

Focus on the negative xx-axis → opens left → form y2=−4pxy^2 = -4px

p=2p = 2: y2=−8xy^2 = -8x

Translated Parabola

(x−h)2=4p(y−k)(x-h)^2 = 4p(y-k)

opens up/down, vertex at (h,k)(h, k).

(y−k)2=4p(x−h)(y-k)^2 = 4p(x-h)

opens right/left, vertex at (h,k)(h, k).

🔦 The Reflective Property

Parabolas have a remarkable property: any ray parallel to the axis reflects off the parabola and passes through the focus.

Applications

  • Satellite dishes: Incoming parallel signals reflect to the focus (receiver)
  • Car headlights: Light placed at the focus reflects outward in parallel beams
  • Solar concentrators: Parallel sunlight focuses to a single point
  • Suspension bridges: Cables under uniform load form parabolas

The focal length pp determines how "wide" or "narrow" the parabola opens. A small pp creates a narrow, tightly focused parabola.

Parabola Quiz 🎯

Parabola Calculations 🧮

1) For x2=16yx^2 = 16y, find pp: p = ?

2) For y2=−24xy^2 = -24x, the focus is at (−a,0)(-a, 0). What is aa?

3) A parabola opens upward with focus at (0,5)(0, 5). Its equation is x2=?yx^2 = ?y. Enter the coefficient.

Parabola Properties 🔽

Exit Quiz ✅

Part 4: Hyperbolas

📐 The Hyperbola

Part 4 of 7

Definition

A hyperbola is the set of all points where the difference of distances to two foci is constant: ∣d1−d2∣=2a|d_1 - d_2| = 2a.

Standard Forms (Center at Origin)

OpensEquationVerticesFociAsymptotes
Left-Rightx2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(±a,0)(\pm a, 0)(±c,0)(\pm c, 0)y=±baxy = \pm\frac{b}{a}x
Up-Downy2a2−x2b2=1\frac{y^2}{a^2}-\frac{x^2}{b^2}=1(0,±a)(0, \pm a)(0,±c)(0, \pm c)y=±abxy = \pm\frac{a}{b}x

Key relationship: c2=a2+b2c^2 = a^2 + b^2 (note the + sign, unlike the ellipse!)

📝 Worked Example

Analyze x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1

a2=9,b2=16  ⟹  a=3,b=4a^2 = 9, b^2 = 16 \implies a = 3, b = 4

c2=9+16=25  ⟹  c=5c^2 = 9 + 16 = 25 \implies c = 5

  • Center: (0,0)(0, 0)
  • Vertices: (±3,0)(\pm 3, 0)
  • Foci: (±5,0)(\pm 5, 0)
  • Asymptotes: y=±43xy = \pm\frac{4}{3}x
  • Eccentricity: e=ca=53≈1.67e = \frac{c}{a} = \frac{5}{3} \approx 1.67

Graphing Strategy

  1. Plot vertices at (±a,0)(\pm a, 0)
  2. Mark co-vertices at (0,±b)(0, \pm b)
  3. Draw the central rectangle through these four points
  4. Draw asymptotes as diagonals of this rectangle
  5. Sketch the hyperbola approaching but never touching the asymptotes

🧠 Hyperbola vs. Ellipse

PropertyEllipseHyperbola
Definitiond1+d2=2ad_1 + d_2 = 2a$
cc relationshipc2=a2−b2c^2 = a^2 - b^2c2=a2+b2c^2 = a^2 + b^2
ShapeClosed curveTwo open branches
Eccentricity0<e<10 < e < 1e>1e > 1
AsymptotesNoney=±baxy = \pm\frac{b}{a}x

Eccentricity of Hyperbolas

  • ee close to 11: branches open very narrowly
  • e=2e = \sqrt{2}: rectangular hyperbola (a=ba = b)
  • ee large: branches open very widely

💡 The conjugate axis has length 2b2b; the transverse axis has length 2a2a and connects the vertices.

Hyperbola Quiz 🎯

Hyperbola Calculations 🧮

For y236−x264=1\frac{y^2}{36}-\frac{x^2}{64}=1:

1) aa = ?

2) cc = ?

3) Eccentricity ee = ? (Enter as a fraction)

Hyperbola Properties 🔽

Exit Quiz ✅

Part 5: Identifying Conics

🔄 Rotated Conics & General Second-Degree Equations

Part 5 of 7

The General Second-Degree Equation

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

When B≠0B \neq 0, the conic is rotated — its axes are not aligned with the coordinate axes.

The Discriminant Test (Revisited)

Δ=B2−4AC\Delta = B^2 - 4AC determines the type:

DiscriminantConic
Δ<0\Delta < 0Ellipse (or circle if A=C,B=0A=C, B=0)
Δ=0\Delta = 0Parabola
Δ>0\Delta > 0Hyperbola

💡 The discriminant is invariant under rotation — changing coordinates doesn't change B2−4ACB^2-4AC.

🔀 Eliminating the xyxy-Term

To remove the BxyBxy term, rotate axes by angle θ\theta where:

cot⁡2θ=A−CB\cot 2\theta = \frac{A-C}{B}

Rotation Formulas

x=Xcos⁡θ−Ysin⁡θx = X\cos\theta - Y\sin\theta y=Xsin⁡θ+Ycos⁡θy = X\sin\theta + Y\cos\theta

Substituting transforms Ax2+Bxy+Cy2+…=0Ax^2+Bxy+Cy^2+\ldots = 0 into:

A′X2+C′Y2+D′X+E′Y+F′=0A'X^2 + C'Y^2 + D'X + E'Y + F' = 0

with no XYXY-term — now in standard position!

Example

xy=1xy = 1 has A=0,B=1,C=0A=0, B=1, C=0. cot⁡2θ=0  ⟹  θ=45°\cot 2\theta = 0 \implies \theta = 45°.

After rotation: X22−Y22=1\frac{X^2}{2} - \frac{Y^2}{2} = 1 — a hyperbola!

🏷️ Classification Practice

Example 1: x2+4xy+4y2+2x−3y+1=0x^2 + 4xy + 4y^2 + 2x - 3y + 1 = 0

A=1,B=4,C=4A=1, B=4, C=4

Δ=16−16=0\Delta = 16 - 16 = 0 → Parabola

Example 2: 2x2+3xy−2y2+x−5=02x^2 + 3xy - 2y^2 + x - 5 = 0

A=2,B=3,C=−2A=2, B=3, C=-2

Δ=9−4(2)(−2)=9+16=25>0\Delta = 9 - 4(2)(-2) = 9+16 = 25 > 0 → Hyperbola

Example 3: 3x2+2xy+3y2−8=03x^2 + 2xy + 3y^2 - 8 = 0

A=3,B=2,C=3A=3, B=2, C=3

Δ=4−36=−32<0\Delta = 4 - 36 = -32 < 0 → Ellipse

Note: Degenerate cases (empty set, single point, intersecting lines) can occur when the equation factors.

Classification Quiz 🎯

Discriminant Practice 🧮

Calculate Δ=B2−4AC\Delta = B^2-4AC for each:

1) 5x2+3xy+2y2=105x^2+3xy+2y^2=10: Δ\Delta = ?

2) x2−4xy+4y2+y=0x^2-4xy+4y^2+y=0: Δ\Delta = ?

3) 2x2+5xy−3y2=12x^2+5xy-3y^2=1: Δ\Delta = ?

Rotated Conics Concepts 🔽

Exit Quiz ✅

Part 6: Problem-Solving Workshop

🌍 Applications of Conic Sections

Part 6 of 7

Conics appear throughout science, engineering, and nature:

Planetary Orbits (Ellipses)

Kepler's First Law: planets orbit the Sun in ellipses with the Sun at one focus.

Satellite Dishes & Headlights (Parabolas)

The reflective property of parabolas focuses signals to a single point or projects light in parallel beams.

Hyperbolic Navigation (Hyperbolas)

LORAN (Long Range Navigation) uses the difference in signal arrival times from two stations, which traces a hyperbola.

Architecture (All Conics)

  • Elliptical rooms (whispering galleries)
  • Parabolic arches and bridges
  • Hyperbolic cooling towers

🪐 Orbital Mechanics

A body under gravity follows a conic section. The type depends on its energy:

EnergyOrbit TypeEccentricity
E<0E < 0 (bound)Ellipse0≤e<10 \leq e < 1
E=0E = 0 (escape)Parabolae=1e = 1
E>0E > 0 (unbound)Hyperbolae>1e > 1

Example: Earth's Orbit

a=1.496×108a = 1.496 \times 10^8 km, e=0.0167e = 0.0167

  • Perihelion (closest): a(1−e)=1.471×108a(1-e) = 1.471 \times 10^8 km
  • Aphelion (farthest): a(1+e)=1.521×108a(1+e) = 1.521 \times 10^8 km

The nearly circular orbit (e≈0e \approx 0) gives us relatively stable seasons.

🔊 Acoustic & Optical Applications

Whispering Gallery (Ellipse)

In an elliptical room, sound from one focus reflects off the wall and converges at the other focus. Famous examples:

  • St. Paul's Cathedral, London
  • National Statuary Hall, U.S. Capitol

Parabolic Reflectors

A parabolic mirror reflects all incoming parallel rays to the focus:

  • Telescopes (reflecting telescopes)
  • Solar cookers
  • Microphone dishes (for recording distant sounds)

Hyperbolic Mirrors

Used in Cassegrain telescopes: the secondary mirror is hyperbolic, redirecting light from the primary parabolic mirror to a more convenient focal point.

Applications Quiz 🎯

Application Calculations 🧮

1) A planet orbits with a=10a = 10 AU and e=0.6e = 0.6. Perihelion =a(1−e)== a(1-e) = ? AU

2) Same planet: aphelion =a(1+e)== a(1+e) = ? AU

3) A parabolic dish has equation x2=8yx^2 = 8y (in feet). The receiver should be placed at the focus: y=y = ? feet

Real-World Conics 🔽

Exit Quiz ✅

Part 7: Review & Applications

🎯 Conic Sections — Full Synthesis

Part 7 of 7

Summary of All Conics

ConicEquationeeKey Property
Circlex2+y2=r2x^2+y^2=r^200All points equidistant from center
Ellipsex2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=10<e<10<e<1d1+d2=2ad_1+d_2=2a
Parabolax2=4pyx^2=4py11Equidistant from focus & directrix
Hyperbolax2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1e>1e>1$

The cc-Relationship

  • Ellipse: c2=a2−b2c^2 = a^2 - b^2 (foci inside)
  • Hyperbola: c2=a2+b2c^2 = a^2 + b^2 (foci outside)
  • Parabola: just one focus at distance pp from vertex

🗺️ Identification Strategy

Given a second-degree equation, follow this flowchart:

Step 1: Is there an xyxy-term?

  • Yes → Use discriminant B2−4ACB^2-4AC to classify; rotate if needed
  • No → Go to Step 2

Step 2: Which squared terms are present?

  • Both x2x^2 and y2y^2 → Go to Step 3
  • Only x2x^2 or only y2y^2 → Parabola

Step 3: Are the coefficients of x2x^2 and y2y^2 the same sign?

  • Same sign, same value → Circle
  • Same sign, different values → Ellipse
  • Opposite signs → Hyperbola

Converting to Standard Form

  1. Group xx-terms and yy-terms
  2. Complete the square for each variable
  3. Divide to get 11 on the right side

📝 Comprehensive Example

Identify and graph: 4x2+9y2−16x+54y+61=04x^2+9y^2-16x+54y+61=0

Step 1: Group and complete the square.

4(x2−4x)+9(y2+6y)=−614(x^2-4x) + 9(y^2+6y) = -61

4(x2−4x+4)+9(y2+6y+9)=−61+16+814(x^2-4x+4) + 9(y^2+6y+9) = -61+16+81

4(x−2)2+9(y+3)2=364(x-2)^2 + 9(y+3)^2 = 36

Step 2: Divide by 36.

(x−2)29+(y+3)24=1\frac{(x-2)^2}{9} + \frac{(y+3)^2}{4} = 1

Identify: Ellipse, center (2,−3)(2,-3), a=3a=3, b=2b=2, horizontal major axis.

c=9−4=5c = \sqrt{9-4} = \sqrt{5}, foci at (2±5,−3)(2\pm\sqrt{5}, -3).

Synthesis Quiz 🎯

Complete the Square 🧮

Convert x2+4y2+2x−24y+33=0x^2+4y^2+2x-24y+33=0 to standard form.

1) Center hh = ?

2) Center kk = ?

3) This is a(n): (type "ellipse", "parabola", "hyperbola", or "circle")

Master Classification 🔽

Final Exit Quiz ✅