Hyperbola standard equation

A hyperbola is what you get when, instead of the sum of distances to two fixed points staying constant (that gives an ellipse), the difference stays constant. Pull two pins slightly apart, and instead of tracing a loop, the locus splits into two open curves that shoot off in opposite directions — one branch hugging each pin. This article derives the standard coordinate form of that locus, $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ , starting from the same two-foci definition used for the ellipse, and shows how one sign flip — a plus instead of a minus — changes a closed oval into two runaway branches. Along the way we identify the vertices, foci, and the two axes, and practice going from given vertices and foci straight to the equation.

The core insight

A hyperbola is the locus where the difference of distances to two fixed foci is constant, giving the standard equation $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with $c^2=a^2+b^2$ .

The core idea

Definition. A hyperbola is the set of points $P(x,y)$ such that the difference of distances from two fixed points $F_1$ and $F_2$ , the foci, is constant in magnitude: $|PF_1 - PF_2| = 2a$ . Unlike the ellipse's sum, this is a difference, and that single change in operation is what produces two separate branches instead of one closed loop.

Placing the foci. Put the foci symmetrically on the $x$ -axis, $F_1(-c,0)$ and $F_2(c,0)$ , with the centre at the origin. A point $P(x,y)$ on the locus satisfies

\left|\sqrt{(x+c)^2+y^2} - \sqrt{(x-c)^2+y^2}\right| = 2a,

where $c$ is the distance from the centre to each focus and $2a$ is the constant difference of distances.

Reducing to standard form. Isolating one radical, squaring, simplifying, and squaring again — the same sequence of steps used for the ellipse — reduces this to

\frac{x^2}{a^2} - \frac{y^2}{c^2-a^2} = 1.

For a real hyperbola $c > a$ always (this is the opposite ordering from the ellipse, where $a>c$ ), so $c^2 - a^2$ is a positive number. Writing $b^2 = c^2 - a^2$ gives the standard equation of a hyperbola:

\boxed{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,}

where: $a$ is the semi-transverse axis (half the distance between vertices), $b$ is the semi-conjugate axis defined by $b^2=c^2-a^2$ , and $c$ is the distance from the centre to each focus.

The key relation. Rearranging $b^2 = c^2-a^2$ gives

c^2 = a^2 + b^2.

This plus sign is the single fact that separates a hyperbola from an ellipse (whose relation is $c^2=a^2-b^2$ ). Because it is a sum, $c$ is always larger than both $a$ and $b$ individually could suggest, and — unlike the ellipse — there is no rule that $a$ must exceed $b$ ; either can be the larger.

Parts of the curve. For $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ : the centre is the origin; the transverse axis lies along the $x$ -axis with length $2a$ and vertices at $(\pm a,0)$ — the only points where the curve actually meets that axis; the conjugate axis lies along the $y$ -axis with length $2b$ , marked by $(0,\pm b)$ , which are not points on the curve; the foci sit at $(\pm c,0)$ . Because the equation forces $x^2/a^2 \ge 1$ , no real point exists for $-a<x<a$ , so the locus splits into two branches, one opening right from $(a,0)$ and one opening left from $(-a,0)$ .

Figure

A hyperbola with two branches opening left and right along the x axis, marked vertices at plus or minus a and foci at plus or minus c, with the region between the vertices empty

Derivation

Derivation AWhy the constant difference equals 2a

The goal is to show, using nothing beyond substituting coordinates into the defining property and reading off distances at a single convenient point, that the constant $2a$ appearing in the definition $|PF_1-PF_2|=2a$ is exactly the length of the transverse axis, matching the labelling used for the vertices $(\pm a,0)$ — the same style of check used to confirm the ellipse's constant sum equals its major axis.

Step 1. Take the right vertex $(a,0)$ as the test point, since it lies on the curve and its distances to both foci are easy to compute directly, without needing the full radical equation.

Step 2. Compute the distance from $(a,0)$ to the nearer focus $F_2(c,0)$ : since $c>a$ , this distance is $c-a$ .

Step 3. Compute the distance from $(a,0)$ to the farther focus $F_1(-c,0)$ : this distance is $a-(-c) = a+c$ .

Step 4. Subtract: $PF_1 - PF_2 = (a+c) - (c-a) = 2a$ . This confirms that the constant difference in the definition is exactly $2a$ , the length of the transverse axis — the same value used when we set $a$ as the semi-transverse axis in the standard equation, and directly parallel to how $2a$ was shown to be the sum for the ellipse's major axis.

Worked example

Problem. Find the equation of the hyperbola with vertices $(\pm 3,0)$ and foci $(\pm 5,0)$ , and state its eccentricity.

Step 1. Vertices are $(\pm a, 0)$ , so $a=3 \Rightarrow a^2=9$ .

Step 2. Foci are $(\pm c,0)$ , so $c=5 \Rightarrow c^2=25$ .

Step 3. Use $b^2 = c^2-a^2 = 25-9=16$ .

Step 4. Write the equation: $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ .

Step 5 (check). At the vertex $(3,0)$ , distance to focus $(5,0)$ is $2$ and distance to focus $(-5,0)$ is $8$ ; the difference $8-2=6=2a$ matches $a=3$ , confirming the equation.

Answer: $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ .

Competitive stretch

The eccentricity of a hyperbola is defined as $e=c/a$ , and it is always greater than $1$ (since $c>a$ always). Here $e = 5/3 \approx 1.67$ . A larger $e$ means a "wider," more open hyperbola — the two branches spread apart faster — while an $e$ close to $1$ means a narrow hyperbola hugging its transverse axis. Contrast this with the ellipse, whose eccentricity always lies strictly between $0$ and $1$ .

Common traps

Using $c^2=a^2-b^2$ (the ellipse relation) instead of $c^2=a^2+b^2$ — the plus sign means $c$ is always larger than $a$ , and it means $b$ can be smaller, larger, or equal to $a$ , with no fixed ordering.
Plugging in an $x$ -value with $|x|<a$ and expecting a real $y$ — no points of the curve exist in that strip; the two branches only begin at the vertices.
Confusing the transverse axis (through the vertices, on the curve) with the conjugate axis (through $(0,\pm b)$ , not on the curve, only a reference length).
Deciding which axis the hyperbola opens along by looking at which variable comes first, rather than which squared term is positive — the branches open along the axis of the positive term.

Ask the tutor: Want to see why swapping which term is positive flips the hyperbola from opening left-right to opening up-down, worked through with a concrete example?

Why does a hyperbola have two branches while an ellipse is one closed curve?

It comes straight from the algebra of $x^2/a^2 \ge 1$ . Because the equation forces $x^2 \ge a^2$ , no real $y$ exists for $-a<x<a$ , so the curve cannot connect through the middle — it must split into a branch for $x\ge a$ and a separate one for $x\le -a$ . An ellipse's equation instead bounds $x$ between $-a$ and $a$ with no gap, so it closes into one loop. The root cause is the difference versus sum in each conic's defining property.

Is there any restriction like a is greater than b for a hyperbola, the way there is for an ellipse?

No. For an ellipse, whichever of $a$ or $b$ is larger tells you which axis is major, and the convention fixes $a>b$ . A hyperbola has no such rule: $b$ can be smaller than, larger than, or equal to $a$ , because $b^2=c^2-a^2$ only requires $c>a$ , and $c$ is unrelated to how $b$ compares with $a$ . What matters for orientation is only which squared term (the one under $x^2$ or under $y^2$ ) carries the positive sign.

How do I tell if a given equation is a hyperbola opening left-right or up-down?

Look only at which squared term is positive. In $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ , the $x^2$ term is positive, so the branches open left and right along the $x$ -axis, with vertices $(\pm a,0)$ . In $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ , the $y^2$ term is positive instead, so the branches open up and down along the $y$ -axis, with vertices $(0,\pm a)$ . The denominator under the positive term is always $a^2$ , giving the transverse axis and the vertices; do not assume $x^2$ 's denominator is automatically $a^2$ .