Ellipse axes and focus relationship

Take the standard ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b>0$, so the major axis lies along the $x$-axis.

• The semi-major axis $a$ is half the longer axis. The vertices sit at $(\pm a,0)$ and the full major axis has length $2a$.
• The semi-minor axis $b$ is half the shorter axis, perpendicular to the major one. It ends at $(0,\pm b)$, and the full minor axis has length $2b$.
• The focal distance $c$ is how far each focus sits from the centre, with the two foci at $(\pm c,0)$.

These three are locked together:

where: $a$ is the semi-major axis, $b$ is the semi-minor axis, and $c$ is the distance from the centre to a focus, all measured in the same length unit. This is not an extra fact bolted on to the ellipse — it is baked into the equation. When the ellipse's equation is first derived from the constant-sum property $PF_1+PF_2=2a$, the quantity $b^2$ is defined as $a^2-c^2$ purely to make $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ come out in that clean form. Because $c^2=a^2-b^2\ge 0$, it follows immediately that $a\ge b$ always; if $c=0$ the two foci merge at the centre and the ellipse becomes a circle. Push $c$ toward $0$ and the ellipse looks rounder; push $c$ toward $a$ and it flattens into a thin sliver, since $b\to 0$ in that limit. If instead the larger denominator sits under $y^2$ (so $b>a$), the major axis runs along the $y$-axis, the foci sit at $(0,\pm c)$, and the relation flips to $c^2=b^2-a^2$ — same idea, larger semi-axis renamed. It helps to think of $a$, $b$ and $c$ as three lengths tied by one constraint rather than three free numbers: fixing any two pins down the third, and every question in this topic is really just asking you to run that one relation in whichever direction the given information demands.

Look at the point $B(0,b)$, the top of the minor axis, together with the two foci $F_1(-c,0)$ and $F_2(c,0)$. By the symmetry of the ellipse, $BF_1=BF_2$, and this pair of equal focal distances together with the triangle $F_1OB$ (right-angled at the centre $O$) is exactly what the derivation below turns into the boxed relation.

Problem. An ellipse has semi-major axis $13$ and semi-minor axis $5$, with its major axis along the $x$-axis. Find the coordinates of its foci, and verify the result using the right triangle.

Step 1. Identify the given lengths: $a=13$, $b=5$.

Step 2. Apply the relation: $c^2=a^2-b^2=169-25=144$.

Step 3. Take the square root: $c=\sqrt{144}=12$.

Step 4. Since the major axis is along the $x$-axis, the foci lie at $(\pm c,0)$. Answer: foci at $(12,0)$ and $(-12,0)$.

Step 5. Check via Pythagoras: $b^2+c^2=25+144=169=13^2=a^2$, confirming the right-triangle relation holds for these values.

Competitive stretch

An ellipse with major axis along the $y$-axis has foci at $(0,\pm 6)$ and semi-major axis $10$. Here the roles swap: the semi-major length is $10$ (paired with $y^2$) and $c=6$, so $b^2=10^2-6^2=100-36=64$, giving $b=8$. The equation is $\dfrac{x^2}{64}+\dfrac{y^2}{100}=1$, and the check $b^2+c^2=64+36=100=10^2$ confirms it again — the same relation, applied with $a$ and $b$'s geometric roles (not their algebraic denominators) tracked carefully. Notice that both worked examples used the identical three-step chain — read off two of $a$, $b$, $c$ from the question, apply $c^2=a^2-b^2$ in whichever direction is needed, and verify with the Pythagoras check — which is exactly the pattern to drill until it is automatic.

• Writing $c^2=a^2+b^2$, the hyperbola relation, instead of $c^2=a^2-b^2$ for the ellipse — the sign is the single most common slip between the two conics.
• Forgetting that $a$ always denotes the larger of the two axis semi-lengths, regardless of whether it is paired with $x^2$ or $y^2$ in the equation.
• Assuming $c$ can equal or exceed $a$ — for a real ellipse $0\le c<a$ always, since $c^2=a^2-b^2$ requires $b^2\ge 0$.
• Placing the foci on the minor axis by mistake — the foci always lie on the major axis, never the minor one.
• Skipping the Pythagoras check after computing $c$ — plugging the answer back into $b^2+c^2=a^2$ takes a few seconds and catches almost every arithmetic slip in an exam.

Ask the tutor: Want to see how this same right-triangle relation reappears when we compute an ellipse's eccentricity, or when we compare it to the hyperbola's focus relation?