Hyperbola: Definition, Equation, Foci, and Asymptotes
Hyperbola: Definition, Equation, Foci, and Asymptotes
What Is a Hyperbola?
A hyperbola is the set of all points in a plane for which the difference of the distances to two fixed points, called the foci, is a positive constant. Those two distances need not be equal — only their difference stays fixed, and that constant difference is what bends the curve into two separate branches.
A hyperbola is one of the four conic sections — the curves you get by slicing a cone with a plane. Slice gently and you get a circle or an ellipse; slice parallel to the side and you get a parabola; slice steeply enough to cut both halves of the cone and you get a hyperbola's two branches.
The contrast with the ellipse is the cleanest way to hold the idea. An ellipse is the set of points where the sum of the two focal distances is constant — a closed loop. Swap "sum" for "difference" and the loop breaks open into two mirror curves. That single word, difference versus sum, is the whole distinction.
How Is A Hyperbola Different From A Parabola?
Both curves open outward, so they are easy to confuse. A parabola is a single open curve, defined by one focus and a directrix line; a hyperbola has two branches and two foci. A parabola has no asymptotes; a hyperbola's branches hug two straight asymptote lines as they run off to infinity.
The Standard Equation of a Hyperbola
For a hyperbola centred at the origin with its transverse axis (the axis through both vertices) along the x-axis, the standard equation is:
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
If the transverse axis runs along the y-axis instead, the equation flips:
$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$
The single feature that tells the two apart is which variable carries the positive term. Positive $x^2$ term means the branches open left and right; positive $y^2$ term means they open up and down.
Variable key: $a$ is the distance from the centre to each vertex (so the vertices sit at $(\pm a,0)$ for a horizontal hyperbola); $b$ sets the height of the central rectangle that controls the asymptotes; $c$ is the distance from the centre to each focus.
The focus relation, derived
For a hyperbola, the three lengths are tied together by:
$$c^2 = a^2 + b^2$$
This looks like the ellipse relation but with a plus sign instead of a minus — and the plus is not arbitrary. It falls straight out of the definition. Take the vertex point $(a,0)$, which lies on the curve. Its distance to the near focus is $c-a$ and to the far focus is $c+a$, so the constant difference of focal distances is $(c+a)-(c-a)=2a$. Working that constant through the distance definition for a general point and simplifying produces $c^2 = a^2 + b^2$ — which is why, for a hyperbola, the focus always sits farther from the centre than the vertex.
The Key Parts of a Hyperbola
| Part | For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ |
|---|---|
| Centre | $(0,0)$ |
| Vertices | $(\pm a,0)$ |
| Foci | $(\pm c,0)$, where $c=\sqrt{a^2 + b^2}$ |
| Transverse axis length | $2a$ |
| Conjugate axis length | $2b$ |
| Asymptotes | $y=\pm \frac{b}{a}x$ |
| Eccentricity | $e=\frac{c}{a} = \sqrt{1 + \frac{b^2}{a^2}}$ |
The asymptotes are the two straight lines the branches approach but never touch. The fastest way to draw them: mark the central rectangle of width $2a$ and height $2b$, then extend its diagonals. The eccentricity $e$ measures how "open" the curve is; for every hyperbola $e>1$, and the larger $e$, the wider the branches spread.
Examples of Hyperbola
The examples move from reading values off a standard equation to building one from given parts. Coordinates are in plain units throughout.
Example 1
For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the vertices. Here $a^2=9$, so $a=3$. The positive term is $x^2$, so the hyperbola opens horizontally and the vertices sit on the x-axis:
Vertices: $(\pm 3,0)$
Example 2
For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the foci — and avoid the most common slip. A reader who has just met the ellipse often reaches for $c^2=a^2−b^2$ out of habit, giving $c^2=9−16=-7$. A squared length cannot be negative, so the result is impossible — the formula was the wrong one. That impossibility is the signal to stop. For a hyperbola the relation carries a plus sign:
$$c^2 = a^2 + b^2 = 9 + 16 = 25, \quad c = 5$$ Foci: $(\pm 5,0)$
Example 3
For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the asymptotes. With $a=3$ and $b=4$, the asymptotes are:
y=±43x
Example 4
Find the eccentricity of $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Using $c=5$ from Example 2 and $a=3$:
e=ca=53≈1.67
Example 5
A hyperbola opens up and down with vertices at $(0,±5)$ and foci at $(0,±13)$. Find its equation. The vertices are on the y-axis, so the $y^2$ term is positive and $a=5$. The foci give $c=13$. Solve for $b^2$:
$$b^2 = c^2 - a^2 = 13^2 - 5^2 = 169 - 25 = 144$$
$$\frac{y^2}{25} - \frac{x^2}{144} = 1$$
Example 6
The difference of the distances from a point to two stations at $(±10,0)$ km is always 12 km. Find the hyperbola the point lies on. The constant difference equals $2a$, so $2a=12$ gives $a=6$ and $a^2=36$. The stations are the foci, so $c=10$ and:
$$b^2 = c^2 - a^2 = 100 - 36 = 64$$
$$\frac{x^2}{36} - \frac{y^2}{64} = 1$$
Why the Hyperbola Matters
The hyperbola is the geometry of differences, and differences turn out to run the physical world.
Navigation and positioning — systems that locate a receiver from the difference in signal arrival times place it on a hyperbola; this is the principle behind hyperbolic navigation.
Astronomy — an object passing the Sun with more than escape speed follows a hyperbolic orbit, swinging by once and never returning. Comets on hyperbolic paths are visitors, not residents.
Optics and design — hyperbolic mirrors redirect light toward a focus, a property used in some reflecting telescopes.
The reason the curve keeps appearing is structural: whenever a quantity is governed by a constant difference rather than a constant sum, the geometry is a hyperbola.
Where Hyperbola Problems Go Wrong
Mistake 1: Using the ellipse focus relation
Where it slips in: Right after studying the ellipse, when $c^2=a^2−b^2$ is fresh in memory.
Don't do this: Apply $c^2=a^2−b^2$ to a hyperbola — it can produce a negative number for $c^2$, which is impossible.
The correct way: A hyperbola uses $c^2=a^2+b^2$, because the focus sits farther from the centre than the vertex.
Mistake 2: Misreading which way the hyperbola opens
Where it slips in: Identifying vertices and foci from the equation.
Don't do this: Assume the curve always opens left-right, or read $a$ from whichever denominator is larger.
The correct way: The variable with the positive term sets the opening direction.
Mistake 3: Flipping the asymptote slope
Where it slips in: Writing the asymptotes for a horizontal hyperbola.
Don't do this: Write $y=±abx$ from memory.
The correct way: For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the slopes are $±\frac{b}{a}$ — rise over run from the central rectangle.
Conclusion
A hyperbola is the set of points whose distances to two foci have a constant difference, producing two mirror-image branches.
Its standard equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, with the positive term setting the opening direction.
The focus relation is $c^2 = a^2 + b^2$ — a plus sign, unlike the ellipse.
Asymptotes are $y=±\frac{b}{a}x$, and eccentricity $e=\frac{c}{a} > 1$ for every hyperbola.
The most common mistake is borrowing the ellipse's minus-sign focus relation, which gives an impossible negative $c^2$.
Practice and Next Steps
Work through these problems to solidify your understanding.
Find the vertices, foci, and asymptotes of $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
A hyperbola has vertices $(\pm 4,0)$ and eccentricity $\frac{5}{4}$. Find its equation.
Find the eccentricity of $\frac{y^2}{36} - \frac{x^2}{64} = 1$.