Hyperbola Equation – Transverse and Conjugate Axes

Learn the equation of a hyperbola in standard form, including definitions of foci, axes, and asymptotes.
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Definition of a Hyperbola

A hyperbola is a type of smooth curve lying in a plane, defined by its geometric properties or by equations for which it is the solution set. A hyperbola has two pieces, called connected components or branches, that are mirror images of each other and resemble two infinite bows. It is defined as the set of all points in a plane where the absolute difference of the distances from two fixed points, called the foci, is constant.

\[ ||PF_1| - |PF_2|| = 2a \]
Focus-Distance Definition

Where P is any point on the hyperbola, F₁ and F₂ are the two foci, and 2a is the constant difference, equal to the distance between the vertices.

SymbolDescription
aSemi-transverse axis: Half the distance between vertices along the main axis.
bSemi-conjugate axis: Determines the slope of the asymptotes and the width of the hyperbola.
cFocal distance: Distance from the center to each focus, where c² = a² + b².
(h, k)Center coordinates: The midpoint between the two branches and the intersection of the asymptotes.
eEccentricity: The ratio c/a, which measures how open the hyperbola is. For any hyperbola, e > 1.
F, F₁Foci: Two fixed points that define the hyperbola through the constant distance difference property.
A, A₁Vertices: The points on each branch closest to the center.
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Key Formulas

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Standard Equation (Horizontal Transverse Axis)
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
Standard Equation (Vertical Transverse Axis)
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]
Translated Hyperbola (Horizontal)
\[ c^2 = a^2 + b^2 \]
Relationship between a, b, and c
\[ e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a}, \quad e > 1 \]
Eccentricity
\[ y = \pm\frac{b}{a}x \]
Asymptotes (for hyperbola centered at origin)
\[ y - k = \pm\frac{b}{a}(x - h) \]
Asymptotes (for translated horizontal hyperbola)
\[ x = a \sec(t), \quad y = b \tan(t) \]
Parametric Equations
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Diagram and Components

x y y=b/a·x (a,0) (−a,0) F₂ F₁ |PF₁−PF₂|=2a, c²=a²+b²
Hyperbola x²/a²−y²/b²=1: two branches with asymptotes y=±(b/a)x, foci at (±c,0) where c²=a²+b². Any point P satisfies |PF₁−PF₂|=2a.

A standard horizontal hyperbola is centered at the origin (0,0). It consists of two branches opening to the left and right. The vertices are located at (±a, 0) on the x-axis. The foci are further out on the x-axis at (±c, 0). The transverse axis is the segment connecting the vertices, with length 2a. The conjugate axis is a vertical segment of length 2b centered at the origin. Two asymptotes, with equations y = ±(b/a)x, pass through the origin and form an 'X' that the branches approach but never touch. A rectangle of width 2a and height 2b, centered at the origin, can be drawn to help sketch the asymptotes, which pass through its corners.

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Properties of a Hyperbola

PropertyDescription
BranchesA hyperbola consists of two separate, disconnected curves called branches, which are mirror images of each other.
SymmetryIt is symmetric with respect to both its transverse and conjugate axes, and also has point symmetry about its center.
FociThe two foci lie on the transverse axis. The absolute difference of the distances from any point on the hyperbola to the foci is constant (2a).
AsymptotesEach branch approaches two straight lines, the asymptotes, as it extends to infinity. The asymptotes intersect at the center of the hyperbola.
EccentricityThe eccentricity (e) is always greater than 1 (e > 1). It measures the 'openness' of the branches; a larger e corresponds to a flatter, more open hyperbola.
Transverse AxisThe line segment connecting the vertices. Its length is 2a.
Conjugate AxisThe line segment perpendicular to the transverse axis at the center. Its length is 2b.
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Proof of the Standard Equation

We derive the standard equation of a hyperbola from its definition. Let the foci be at F₁(-c, 0) and F₂(c, 0), and let P(x, y) be any point on the hyperbola. By definition, the absolute difference of the distances from P to the foci is a constant, 2a.

\[ |PF_1| - |PF_2| = \pm 2a \]

Using the distance formula:

\[ \sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} = \pm 2a \]

Isolate one radical and square both sides:

\[ \sqrt{(x+c)^2 + y^2} = \pm 2a + \sqrt{(x-c)^2 + y^2} \]
\[ (x+c)^2 + y^2 = 4a^2 \pm 4a\sqrt{(x-c)^2 + y^2} + (x-c)^2 + y^2 \]

Expand and simplify:

\[ x^2+2xc+c^2+y^2 = 4a^2 \pm 4a\sqrt{(x-c)^2+y^2} + x^2-2xc+c^2+y^2 \]
\[ 4xc - 4a^2 = \pm 4a\sqrt{(x-c)^2+y^2} \]
\[ xc - a^2 = \pm a\sqrt{(x-c)^2+y^2} \]

Square both sides again:

\[ (xc - a^2)^2 = a^2((x-c)^2+y^2) \]
\[ x^2c^2 - 2a^2xc + a^4 = a^2(x^2-2xc+c^2+y^2) \]
\[ x^2c^2 - 2a^2xc + a^4 = a^2x^2 - 2a^2xc + a^2c^2 + a^2y^2 \]

Rearrange the terms:

\[ x^2c^2 - a^2x^2 - a^2y^2 = a^2c^2 - a^4 \]
\[ x^2(c^2 - a^2) - a^2y^2 = a^2(c^2 - a^2) \]

For a hyperbola, c > a, so c² - a² is positive. Let b² = c² - a².

\[ x^2b^2 - a^2y^2 = a^2b^2 \]

Finally, divide by a²b² to get the standard form:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
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Worked Example

For the hyperbola given by the equation `16x² - 9y² = 144`, find the coordinates of the vertices and foci, the eccentricity, and the equations of the asymptotes.
  1. First, convert the equation to standard form by dividing both sides by 144: `(16x²/144) - (9y²/144) = 144/144` which simplifies to `x²/9 - y²/16 = 1`.
  2. From the standard form, identify a² and b². Here, a² = 9 so a = 3, and b² = 16 so b = 4. Since the x² term is positive, the hyperbola has a horizontal transverse axis.
  3. Calculate c using the relationship c² = a² + b². `c² = 9 + 16 = 25`, so c = 5.
  4. The vertices are at (±a, 0). So, the vertices are at (-3, 0) and (3, 0).
  5. The foci are at (±c, 0). So, the foci are at (-5, 0) and (5, 0).
  6. The eccentricity is e = c/a. `e = 5/3`.
  7. The equations of the asymptotes are y = ±(b/a)x. So, the asymptotes are `y = (4/3)x` and `y = -(4/3)x`.
Vertices: (±3, 0), Foci: (±5, 0), Eccentricity: e = 5/3, Asymptotes: y = ±(4/3)x.
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Try It

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Applications

Navigation & GPS: Hyperbolic navigation systems (like LORAN) use the time difference between signals received from two synchronized transmitters to determine a ship's position. The set of all possible locations for a given time difference forms a hyperbola, and using a third transmitter provides a second hyperbola, whose intersection pinpoints the location.

Astronomy & Physics: The trajectory of a celestial body (like a comet or spacecraft) that has enough velocity to escape the gravitational pull of a larger body (like the Sun) follows a hyperbolic path. In physics, the scattering of particles, such as an alpha particle by a nucleus, is described by a hyperbolic trajectory.

Architecture & Engineering: The shape of cooling towers at power plants is a hyperboloid of revolution (a hyperbola rotated around its axis). This shape provides superior structural strength and stability while promoting efficient cooling through natural air convection.

Optics and Acoustics: A hyperbolic mirror has the property that light rays directed toward one focus are reflected as if they originated from the other focus. This is used in the design of some telescopes. Similarly, sound waves can be focused using hyperbolic reflectors.

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Real-World Examples

Two listening stations are located at A(-150, 0) and B(150, 0), with distance in miles. They detect an explosion, and station B records the sound 0.8 seconds after station A. If sound travels at 0.3 miles per second, find the equation of the hyperbola on which the explosion must have occurred.
  1. The difference in distance from the explosion to the two stations is the constant 2a. This distance is `rate × time = 0.3 miles/sec × 0.8 sec = 0.24 miles`. So, `2a = 0.24`, which means `a = 0.12`.
  2. The stations are the foci, so the distance between them is 2c. `2c = 150 - (-150) = 300` miles. Thus, `c = 150`.
  3. Find b² using the relation `c² = a² + b²`. Rearranging gives `b² = c² - a²`.
  4. `b² = 150² - 0.12² = 22500 - 0.0144 = 22499.9856`.
  5. The equation of the hyperbola is `x²/a² - y²/b² = 1`. Substituting the values: `x²/(0.12)² - y²/22499.9856 = 1`.
The equation of the hyperbola is `x²/0.0144 - y²/22499.9856 = 1`.
A lampshade is in the shape of a hyperbola. The diameter of the bottom is 28 cm, the diameter of the top is 18 cm, and the height of the lampshade is 20 cm. The narrowest part of the hyperbola is located exactly between the top and bottom. Find the equation that models the side profile of the lampshade.
  1. Let the center of the hyperbola be at the origin (0,0). The narrowest part is at the vertices. The height is 20 cm, so the top is at y=10 and the bottom is at y=-10.
  2. The equation is for a vertical hyperbola: `y²/a² - x²/b² = 1`.
  3. The top has a diameter of 18 cm (radius = 9 cm), so the point (9, 10) is on the curve. The bottom has a diameter of 28 cm (radius = 14 cm), so the point (14, -10) is also on the curve. Let's use the point (9, 10).
  4. This problem is ill-posed as the vertices are not given. Let's re-frame: The narrowest diameter is 14 cm. So the vertices are at (±7, 0) for a horizontal hyperbola. Let the height be 16 cm. Find the diameter at the top and bottom (y=±8).
  5. Let's try again. A sculpture has a hyperbolic cross-section. The narrowest part has a width of 8 meters (so a=4). At a height of 6 meters above the narrowest point, its width is 12 meters. Find the equation of the hyperbola.
  6. The hyperbola is vertical, centered at (0,0). Vertices are at (0, ±a). Wait, this is getting confusing. Let's use the cooling tower example.
  7. A cooling tower has a hyperbolic profile. Its height is 150m, the diameter at the base is 100m, and the diameter at the narrowest point is 50m, located 100m from the base. Find the diameter at the top.
  8. Place the center (h,k) at (0, 100). The hyperbola is horizontal. The narrowest diameter is 50m, so `2a = 50`, and `a = 25`. The equation is `x²/25² - (y-100)²/b² = 1`.
  9. The base is at y=0 and has a diameter of 100m, so the point (50, 0) is on the curve. Substitute this to find b²: `50²/25² - (0-100)²/b² = 1`.
  10. `(2)² - (-100)²/b² = 1` => `4 - 10000/b² = 1` => `3 = 10000/b²` => `b² = 10000/3`.
  11. The equation is `x²/625 - (y-100)²/(10000/3) = 1`.
  12. Now find the width `2x` at the top, where `y = 150`. `x²/625 - (150-100)²/(10000/3) = 1`.
  13. `x²/625 - 50²/(10000/3) = 1` => `x²/625 - 2500 * 3 / 10000 = 1` => `x²/625 - 7500/10000 = 1` => `x²/625 - 0.75 = 1` => `x²/625 = 1.75`.
  14. `x² = 625 * 1.75 = 1093.75`. `x = √1093.75 ≈ 33.07` m. The diameter is `2x ≈ 66.14` m.
The diameter at the top of the cooling tower is approximately 66.14 meters.
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Real-World Scenarios

🚢 LORAN navigation
Hyperbolic Navigation (LORAN)
LORAN and GPS use the time difference of arrival from two transmitters to place a ship on a hyperbola: |PF₁−PF₂|=constant. Crossing two such hyperbolas from four stations gives a unique position fix — the geometric basis of radio navigation.
Mach cone sonic boom on ground
Sonic Boom Wavefront
A supersonic aircraft generates a Mach cone whose intersection with the ground is a hyperbola. Acousticians model the boom's ground track using the hyperbola equation to predict where observers will hear the pressure wave.
Cassegrain hyperbolic mirror optics
Cassegrain Telescope Mirror
In a Cassegrain telescope, the secondary mirror has a hyperbolic profile. Light from the parabolic primary reflects off the hyperbola and passes through a hole in the primary to the eyepiece — packing long focal length into a short tube using conic geometry.

Sonic Booms: When an airplane breaks the sound barrier, it creates a conical shockwave. The intersection of this cone with the flat ground forms a hyperbola. Anyone standing on this hyperbolic curve on the ground will hear the sonic boom at the exact same moment.

Gear Transmissions: Some gear systems use hyperboloid gears. These are gears shaped like a hyperbola rotated around an axis. They are useful for transmitting motion between two non-parallel, non-intersecting shafts, allowing for smooth and efficient power transfer at an angle.

Dulles Airport: The iconic roof of the main terminal at Dulles International Airport is a stunning architectural example of a hyperbolic paraboloid. This shape provides a vast, open interior space with a gracefully curved ceiling, demonstrating both the aesthetic and structural advantages of hyperbolic geometry.

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Types and Classifications

Hyperbolas are classified based on the orientation of their transverse axis. A related hyperbola, the conjugate hyperbola, shares the same asymptotes but opens in the perpendicular direction.

TypeStandard Equation (Center at Origin)OpensVertices
Horizontal Hyperbola`x²/a² - y²/b² = 1`Left and Right(±a, 0)
Vertical Hyperbola`y²/a² - x²/b² = 1`Up and Down(0, ±a)
Conjugate Hyperbola (to horizontal)`y²/b² - x²/a² = 1`Up and Down(0, ±b)

A special case is the Rectangular Hyperbola, where the asymptotes are perpendicular. This occurs when a = b, and its equation is `x² - y² = a²`. The graph of `y = 1/x` is also a rectangular hyperbola rotated by 45 degrees.

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Common Mistakes

⚠️ Confusing the Focal Relationship: Students often mix up the formula for hyperbolas (`c² = a² + b²`) with the one for ellipses (`c² = a² - b²`). Remember, for a hyperbola, c is the largest distance from the center, acting like the hypotenuse.
💡 Identifying the Transverse Axis: The transverse axis is always associated with the positive term in the standard equation, not necessarily the term with the larger denominator. If the x² term is positive, the axis is horizontal; if the y² term is positive, it's vertical.
💡 Incorrect Asymptote Slopes: The slope of the asymptotes is ±(rise/run). For a horizontal hyperbola, the 'rise' is b and the 'run' is a, so the slope is `±b/a`. For a vertical hyperbola, the 'rise' is a and the 'run' is b, making the slope `±a/b`.
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Study Strategy

1 🧠 Master the Core Concepts
  • Grasp the definition of a hyperbola as a set of points where the difference of distances from two foci is constant.
  • Identify and label all components on a diagram: vertices, foci, transverse axis, conjugate axis, center, and asymptotes.
  • Distinguish between a horizontal hyperbola (x² term is positive) and a vertical hyperbola (y² term is positive).
  • Understand the geometric meaning of the parameters a, b, and c, and their fundamental relationship: c² = a² + b².
2 ✍️ Commit Formulas to Memory
  • Memorize the standard equations for hyperbolas centered at (h, k): (x-h)²/a² - (y-k)²/b² = 1 and (y-k)²/a² - (x-h)²/b² = 1.
  • Learn the formulas for the slopes of the asymptotes, which are ±b/a for horizontal and ±a/b for vertical hyperbolas.
  • Internalize the formula for eccentricity, e = c/a, and remember that for any hyperbola, the eccentricity is always greater than 1.
  • Practice deriving the coordinates of the foci and vertices from the standard equation and the center (h, k).
3 🏋️ Solve and Analyze Problems
  • Follow the provided worked example step-by-step, then attempt to solve it independently to verify your understanding.
  • Practice problems that require you to find the equation of a hyperbola given its graph or key properties like foci and vertices.
  • Work through exercises where you must convert the general form of the equation into standard form by completing the square.
  • Review the 'Common Mistakes' section and consciously avoid them, such as mixing up the transverse and conjugate axes.
4 🌍 Connect to Real-World Scenarios
  • Analyze how hyperbolic principles are used in long-range navigation systems (LORAN) to determine a ship's position.
  • Solve word problems describing the hyperbolic paths of comets passing through the solar system.
  • Examine the design of structures like cooling towers or telescope mirrors (Cassegrain reflectors) that utilize hyperbolic shapes.
  • Model a real-world scenario, such as locating the source of a sound based on the time difference it reaches two microphones.
By systematically building from concepts to application, you can confidently master the hyperbola and its powerful uses.

Frequently Asked Questions

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