Ellipse

Pin a loop of string at two fixed points, pull it taut with a pencil, and trace around — the curve you draw is an ellipse, the locus of every point whose distances from those two pins add up to the same constant total. It is the shape of planetary orbits, whispering galleries, and the "squashed circle" you see when a circular plate is viewed at a slant. In ISC Class 11 this sits in the optional Section B "Conic Section" unit, alongside the circle and the parabola, and it rewards the same skill: read off centre, vertices, foci, eccentricity, and latus rectum directly from a standard equation, or build that equation from given data. The algebra is short — one Pythagoras-like relation, $b^2 = a^2 - c^2$ , does almost all the work — but the geometry it encodes (two foci, a constant sum, an oval that is never a circle unless $a = b$ ) is worth holding in your head as a picture, not just a formula.

The core insight

An ellipse is the set of points whose distances from two fixed foci sum to a constant $2a$ , giving the standard equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $b^2=a^2-c^2$ and eccentricity $e=c/a\in(0,1)$ .

The core idea

Place the centre of the ellipse at the origin with the two foci on the $x$ -axis at $(\pm c, 0)$ . A point $P(x,y)$ lies on the ellipse exactly when the sum of its distances to the two foci is a fixed constant, written $2a$ for convenience:

PF_1 + PF_2 = 2a, \qquad a > c > 0

where: $F_1(-c,0)$ and $F_2(c,0)$ are the foci, and $2a$ is the constant total distance (the length of string in the pin-and-loop picture). Squaring twice to clear the two square roots in $PF_1$ and $PF_2$ turns this distance condition into the familiar algebraic form.

Standard equation. With $b^2 = a^2 - c^2$ (so $a > b > 0$ ), the locus becomes

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

where: $a$ is the semi-major axis (half the longer axis, here along $x$ ), $b$ is the semi-minor axis (half the shorter axis, here along $y$ ), and $c = \sqrt{a^2-b^2}$ is the distance from centre to each focus.

Reading off the features. For this orientation ( $a > b$ , major axis along $x$ ):

Centre $(0,0)$ ; vertices (ends of major axis) $(\pm a, 0)$ ; ends of minor axis $(0,\pm b)$ .
Foci $(\pm c, 0)$ with $c=\sqrt{a^2-b^2}$ .
Eccentricity $e = \dfrac{c}{a}$ , always satisfying $0 < e < 1$ for a genuine ellipse — a value near $0$ is nearly circular, a value near $1$ is very flattened.
Latus rectum, the focal chord perpendicular to the major axis, has length $\dfrac{2b^2}{a}$ .

If instead the larger denominator sits under $y^2$ (major axis along $y$ ), every feature simply swaps axis: vertices and foci lie on the $y$ -axis, and $e$ uses the larger semi-axis in its denominator. The string picture also explains why $a$ must be the larger semi-axis: the loop has to reach past both foci and back, so its half-length $a$ must exceed the half-distance between foci, $c$ — precisely the condition $a>c$ used above, and the reason $e=c/a$ is always below $1$ .

Why eccentricity measures shape. Because $b^2=a^2-c^2$ , the two extreme cases are easy to picture. As $c\to0$ the two foci merge into the centre, $b\to a$ , and the ellipse becomes a circle — its most "round" possible shape, with $e\to0$ . As $c\to a$ the foci separate almost to the vertices, $b\to0$ , and the ellipse flattens into a near-line-segment, with $e\to1$ . Every real ellipse sits strictly between these extremes, so a quick glance at $e$ tells you at once whether the curve is nearly circular (small $e$ , planetary orbits are typically close to this) or noticeably elongated (larger $e$ , comets can approach $e$ close to $1$ while still remaining bound ellipses).

Figure

An ellipse centred at the origin with major axis along x, showing the two foci inside the curve, the semi-major and semi-minor axes marked, and a point on the curve joined to both foci by line segments whose lengths sum to a constant

Derivation AThe relation between semi-major axis, semi-minor axis and focal distance

The goal is to show algebraically why the ellipse with foci $(\pm c,0)$ and constant distance-sum $2a$ collapses to the tidy equation $x^2/a^2 + y^2/b^2 = 1$ with $b^2 = a^2-c^2$ , so that every later feature — vertices, foci, eccentricity, latus rectum — can be read straight off the two numbers $a$ and $b$ without redoing the distance argument each time.

Step 1. Write the distance-sum condition for a point $P(x,y)$ with foci $F_1(-c,0)$ and $F_2(c,0)$ : $\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2a$ .

Step 2. Isolate one square root and square once: $\sqrt{(x+c)^2+y^2} = 2a - \sqrt{(x-c)^2+y^2}$ . Squaring both sides and expanding removes the left root, and after cancelling the common $x^2+y^2+c^2$ terms on both sides this simplifies to $a\sqrt{(x-c)^2+y^2} = a^2 - cx$ .

Step 3. Square once more to remove the remaining root: $a^2\big[(x-c)^2+y^2\big] = (a^2-cx)^2$ . Expanding both sides and cancelling the common terms $a^2x^2$ and $a^2c^2$ leaves $(a^2-c^2)x^2 + a^2y^2 = a^2(a^2-c^2)$ .

Step 4. Name $b^2 = a^2 - c^2$ (positive since $a>c$ ) and divide through by $a^2b^2$ : the equation becomes exactly $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ . This single substitution is what lets every ellipse feature — foci at $c=\sqrt{a^2-b^2}$ , eccentricity $e=c/a$ , latus rectum $2b^2/a$ — be read off from $a$ and $b$ alone, with no need to return to the original distance-sum definition.

Worked example

Find the vertices, foci, eccentricity, and latus rectum of $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ , then find the equation of a second ellipse with major axis length $20$ along the $x$ -axis and eccentricity $\dfrac{3}{5}$ .

Step 1. For $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ : match to $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , giving $a^2=25\Rightarrow a=5$ and $b^2=9\Rightarrow b=3$ . Since $a>b$ , the major axis runs along $x$ .

Step 2. Find $c$ : $c=\sqrt{a^2-b^2}=\sqrt{25-9}=\sqrt{16}=4$ . So vertices are $(\pm5,0)$ and foci are $(\pm4,0)$ .

Step 3. Eccentricity and latus rectum: $e=\dfrac{c}{a}=\dfrac{4}{5}=0.8$ (correctly inside $0<e<1$ ), and latus rectum $=\dfrac{2b^2}{a}=\dfrac{2(9)}{5}=\dfrac{18}{5}=3.6$ .

Step 4. Now the second ellipse: major axis length $2a=20\Rightarrow a=10$ . Eccentricity $e=c/a=3/5$ gives $c=\dfrac{3}{5}(10)=6$ . Then $b^2=a^2-c^2=100-36=64$ , so $b=8$ .

Answer. First ellipse: vertices $(\pm5,0)$ , foci $(\pm4,0)$ , $e=0.8$ , latus rectum $3.6$ . Second ellipse: $\mathbf{\dfrac{x^2}{100}+\dfrac{y^2}{64}=1}$ , checked by $c=\sqrt{100-64}=6$ matching the value used.

Common traps

Assuming the larger denominator always sits under $x^2$ . Whichever of $a^2,b^2$ is larger fixes the major axis — if $b>a$ in $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , the major axis and foci are actually on the $y$ -axis.
Mixing up $b^2=a^2-c^2$ with unrelated conic relations, such as the parabola's $y^2=4ax$ — this relation is specific to the ellipse (with a sign flip, the hyperbola).
Forgetting that eccentricity must satisfy $0<e<1$ . A computed value at or above $1$ signals an arithmetic slip, not a valid ellipse.
Confusing the latus rectum $\big(2b^2/a\big)$ with the minor axis length $(2b)$ — the latus rectum is a specific focal chord, generally shorter than the minor axis.

Ask the tutor: If you're unsure whether your ellipse's major axis runs along $x$ or $y$ from a given equation, ask the AI tutor to check your identification of $a$ , $b$ , and the foci step by step.

Why is a always the semi-major axis and not just whichever letter I choose?

Because $a$ is defined by the geometry, not by which symbol looks bigger in an equation. The string-and-pins picture requires the loop length $2a$ to be long enough to stretch past both foci and back, which forces $a>c$ and, through $b^2=a^2-c^2$ , also forces $a>b$ . So whichever denominator in $\dfrac{x^2}{p}+\dfrac{y^2}{q}=1$ is numerically larger is always the true $a^2$ , regardless of whether it sits under $x^2$ or $y^2$ — check the numbers, not the position.

Why does eccentricity have to be between 0 and 1 for an ellipse?

Eccentricity is $e=c/a$ , and the ellipse's defining condition $a>c>0$ (the loop must be longer than the gap between foci, but the foci must be at some positive distance apart to form a genuine ellipse rather than a circle) forces $0<e<1$ . As $e\to0$ , $c\to0$ and the ellipse becomes a circle; as $e\to1$ , the ellipse flattens toward a line segment. Eccentricities of $0$ or $1$ are the boundary cases, not true ellipses.

How is the ellipse's b squared equals a squared minus c squared different from the hyperbola's similar-looking relation?

For an ellipse the two axes and the focal distance satisfy $b^2=a^2-c^2$ , because $c<a$ — the foci sit strictly inside the curve. For a hyperbola the corresponding relation is $b^2=c^2-a^2$ , because there $c>a$ — the foci sit outside the two branches. The sign flip reflects the different geometry: an ellipse's distance-sum condition versus a hyperbola's distance-difference condition. Keep the two formulas mentally tagged to their curve so you don't swap the sign under exam pressure.

Ellipse

The core idea

Worked example

Common traps

Before this

Where this leads