Hyperbola eccentricity

Setting up the definition. For the standard hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, the foci lie at $(\pm c,0)$ on the transverse axis, a distance $c$ from the centre. The eccentricity is defined exactly as it is for the ellipse — as the ratio of focal distance to semi-transverse-axis length:

where: $c$ is the distance from the centre to a focus, and $a$ is the semi-transverse axis (half the distance between the two vertices).

From $a$ and $b$ instead of $c$. The hyperbola's own axis relation is $c^2=a^2+b^2$ — note the plus sign, the single detail that separates a hyperbola from an ellipse (whose relation is $b^2=a^2-c^2$). Substituting $c=\sqrt{a^2+b^2}$ into $e=c/a$ gives the working formula every problem actually uses:

where: $a$ is the semi-transverse axis, $b$ is the semi-conjugate axis, and $e$ comes out dimensionless, as any ratio of lengths must.

Why $e$ can never equal or dip below $1$. A hyperbola only exists when $b^2>0$ — a zero conjugate axis would collapse the curve to a pair of rays, not a hyperbola. So $b^2/a^2$ is always a strictly positive number, which forces $e=\sqrt{1+b^2/a^2}$ to be strictly greater than $\sqrt{1}=1$, regardless of how small or large $a$ and $b$ are individually. There is no matching upper bound: making $b$ far larger than $a$ pushes $e$ as high as you like, with the branches flattening out toward a pair of straight lines in the limit.

Reading $e$ geometrically. A value of $e$ close to $1$ (small $b$ relative to $a$) describes a narrow hyperbola whose branches hug the transverse axis closely. A large $e$ (large $b$ relative to $a$) describes a hyperbola whose branches splay open widely, almost like two intersecting lines. This same $e$ also appears in the focus–directrix description of the hyperbola: for any point $P$ on the curve, the ratio of its distance from a focus to its distance from the corresponding directrix equals $e$, and because $e>1$ here, a point on a hyperbola is always farther from the focus than from the directrix — the opposite ordering from the ellipse.

One unifying scale for all three conics. Lay the three formulas side by side: ellipse $e=\sqrt{1-b^2/a^2}$ (minus sign, giving $0<e<1$), parabola $e=1$ exactly (a single fixed shape, no $a,b$ needed), hyperbola $e=\sqrt{1+b^2/a^2}$ (plus sign, giving $e>1$ with no ceiling). Every non-circular conic is classified purely by where $e$ sits relative to $1$ — this is why $e$ is often introduced as the one number that unifies the whole conic family.

The rectangular hyperbola, a landmark value. When $a=b$, the equation $\dfrac{x^2}{a^2}-\dfrac{y^2}{a^2}=1$ gives $e=\sqrt{1+1}=\sqrt2$. This tidy, specific number is worth memorising: it is a quick sanity check that $a$ and $b$ have been read off correctly whenever a problem mentions an equilateral (rectangular) hyperbola.

Why the focus lies outside the vertex. Because $c^2=a^2+b^2>a^2$, it follows immediately that $c>a$, which is exactly the statement $e=c/a>1$ written a different way. Geometrically this means each focus always lies farther from the centre than the nearest vertex — outside the "body" of that branch — which is why the branches can curve around and open toward their own focus without ever closing up into a bounded loop the way an ellipse does. This is the qualitative reason a hyperbola has two separate, unbounded branches while an ellipse is a single closed curve: the focus never gets "trapped" inside a branch the way it sits inside an ellipse.

Problem. A hyperbola has transverse axis of length $8$ and eccentricity $e=\dfrac{3}{2}$. Find $b^2$, verify the eccentricity, and state the coordinates of the foci.

Step 1. The transverse axis has length $2a=8$, so $a=4$.

Step 2. From $e=c/a$: $c=ea=\dfrac{3}{2}\times4=6$.

Step 3. Use the axis relation $c^2=a^2+b^2$ to isolate $b^2$: $b^2=c^2-a^2=36-16=20$.

Step 4. Write the full equation using $a^2=16$ and $b^2=20$: $\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$.

Step 5. Verify by recomputing $e$ from the formula in terms of $a$ and $b$: $e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{20}{16}}=\sqrt{\dfrac{36}{16}}=\dfrac{6}{4}=\dfrac{3}{2}$, matching the given value exactly.

Step 6. The foci sit at $(\pm c,0)$.

Answer: $b^2=20$, the hyperbola is $\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$, and the foci are at $(6,0)$ and $(-6,0)$.

Competitive stretch

If instead only $a^2=16$ and $b^2=9$ were given directly (as for $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$), the same two formulas cross-check each other: $e=\sqrt{1+9/16}=\sqrt{25/16}=5/4$ from the $a,b$ formula, and $c=\sqrt{16+9}=5$ gives $e=c/a=5/4$ from the definition — the two routes always agree, which makes this a fast way to catch an arithmetic slip on an exam.

• Writing $e=\sqrt{1-b^2/a^2}$ (the ellipse formula, minus sign) instead of $e=\sqrt{1+b^2/a^2}$ for a hyperbola — the sign is the entire difference between the two formulas.
• Expecting $e<1$ out of habit from the ellipse, and second-guessing a correctly computed value like $e=1.25$ or $e=5/4$ — for a hyperbola, any $e>1$ is valid and there is no upper cap.
• Assuming $a$ and $b$ must obey some size relationship (as with some ellipse conventions) — for a hyperbola $a$ and $b$ can be in any ratio, so a large $b^2/a^2$ and hence a large $e$ is completely normal.
• Sign or arithmetic slips solving $c^2=a^2+b^2$ for an unknown, especially subtracting in the wrong order and landing on a negative $b^2$.
• Forgetting that a rectangular hyperbola ($a=b$) has the specific eccentricity $\sqrt2$, not $1$ or $2$ — worth memorising as a landmark value.

Ask the tutor: Ask the AI tutor to place the ellipse, parabola and hyperbola eccentricity formulas side by side and show how each one traces back to a different focus-and-axis relationship.