Ellipse eccentricity

For the standard ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b>0$ and foci at $(\pm c,0)$, the eccentricity is defined directly as the ratio of the focal distance to the semi-major axis.

where: $c$ is the distance from the centre to a focus, and $a$ is the semi-major axis (the larger of the two denominators' square roots). Because $c$ and $b$ are linked through the axis-focus relation $c^2=a^2-b^2$, the same $e$ can be written purely in terms of the axes, without ever computing $c$ separately.

where: $a$ is the semi-major axis and $b$ is the semi-minor axis, with $a>b>0$ always for the standard form. Both formulas describe the same number; which one you use in a problem just depends on what you are given — the axes $a,b$ (read straight off the equation) or the focal distance $c$ (read off the foci, if they are stated). A few facts about what $e$ is telling you:

• $e$ close to $0$ (so $b$ close to $a$) means the ellipse is nearly circular; $e$ close to $1$ (so $b$ much smaller than $a$) means it is very flattened.
• $e$ depends only on the ratio $b/a$, not on the absolute lengths — scaling an ellipse up or down (doubling both $a$ and $b$) leaves $e$ unchanged, which is exactly why $e$ is called the ellipse's shape "signature."
• Eccentricity also fixes where an ellipse sits among the conics: $e<1$ for an ellipse, $e=1$ for a parabola, $e>1$ for a hyperbola, so measuring $e$ is enough to identify the type of conic.

Eccentricity connects to the more general focus-directrix view of a conic, briefly mentioned when conics are first introduced: for any point $P$ on the curve, the ratio of its distance from a focus to its distance from the corresponding directrix is a constant, and that constant is exactly $e$. For an ellipse this ratio is always less than $1$ — geometrically, points on the ellipse stay closer to the focus than to the directrix, in a fixed proportion — which is the deeper reason $e=1$ marks the parabola and $e>1$ the hyperbola as the ratio crosses over. A familiar (non-syllabus) picture for what $e$ measures: planetary orbits are ellipses with the Sun at one focus, and Earth's orbit has $e\approx0.017$ — so close to $0$ that it looks almost circular — while some comets have $e$ very close to (but still below) $1$, giving them long, thin, elongated paths.

Problem. Find the eccentricity of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$, then use that value of $e$ to find $b$ for a second ellipse that has the same eccentricity but semi-major axis $a=10$.

Step 1. Identify the axes of the first ellipse: the larger denominator is under $x^2$, so $a^2=25$ and $b^2=9$ (confirming $a>b$, as required for the standard form).

Step 2. Apply $e=\sqrt{1-b^2/a^2}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}=0.8$.

Step 3. Cross-check via $c$: $c^2=a^2-b^2=25-9=16\Rightarrow c=4$, so $e=c/a=4/5$, matching Step 2 exactly — the two routes agree, which is a useful habit for catching arithmetic slips.

Step 4. Now use $e=0.8$ on the second ellipse, with $a=10$: from $e=c/a$, $c=ea=0.8\times10=8$. Then $b^2=a^2-c^2=100-64=36\Rightarrow b=6$.

Answer: the first ellipse has $e=\dfrac{4}{5}=0.8$; a second ellipse with the same eccentricity and $a=10$ has $b=6$ (and foci at $(\pm 8,0)$).

Competitive stretch

Problem. An ellipse has semi-major axis $a=10$ and eccentricity $e=0.6$ (a different ellipse from the one above, with a smaller $e$). Find $b$, the coordinates of the foci, and write the ellipse's full standard equation, then verify the eccentricity from the final equation.

Step 1. From $e=c/a$: $c=ea=0.6\times10=6$.

Step 2. Use $c^2=a^2-b^2\Rightarrow b^2=a^2-c^2=100-36=64\Rightarrow b=8$.

Step 3. Foci at $(\pm c,0)=(\pm6,0)$, and the standard equation is $\dfrac{x^2}{100}+\dfrac{y^2}{64}=1$.

Step 4. Verify: $e=\sqrt{1-b^2/a^2}=\sqrt{1-64/100}=\sqrt{36/100}=6/10=0.6$, matching the given value exactly — confirming the ellipse is built correctly. Answer: $b=8$, foci at $(6,0)$ and $(-6,0)$, equation $\dfrac{x^2}{100}+\dfrac{y^2}{64}=1$.

• Thinking higher eccentricity means "rounder" — it is the opposite: $e$ near $0$ is nearly circular, $e$ near $1$ is elongated.
• Using $e=\sqrt{1+b^2/a^2}$ (the hyperbola formula) instead of $e=\sqrt{1-b^2/a^2}$ for the ellipse — the sign is the key difference between the two conics.
• Mixing up $a$ and $b$ before applying the formula — always check which denominator is larger (that one is $a^2$) before computing $e$.
• Getting $e\ge 1$ from a calculation for an ellipse — that always signals an arithmetic or setup error, since a genuine ellipse must have $0<e<1$.
• Forgetting that $e$ alone does not fix the ellipse's size — you also need $a$ (or $c$, or $b$) to pin down an actual equation; $e$ only fixes the shape.

Ask the tutor: Want to see how eccentricity changes an ellipse's shape by comparing several values of $e$ side by side, or how it connects to real orbits?