10.3 The Ellipse
An ellipse is the set of points whose distances from two foci add up to a constant $2a$. With the foci at $(\pm c,0)$ it is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, where $c^2=a^2-b^2$; its eccentricity $e=\dfrac ca$ lies between $0$ and $1$.
Push two drawing pins into a sheet of card, drop a loop of string over them, and pull the loop tight with the tip of a pencil. Now move the pencil all the way round, keeping the string taut. The curve you draw is an ellipse. Gardeners mark out oval flower beds this way with two pegs and a rope. The reason it works is the key to everything in this section: at every moment the two straight pieces of string from the pencil to the pins have the same total length, so every point of the curve has the same sum of distances from the two pins. The pins are the two foci of the ellipse.
Ellipses are everywhere in science. Every planet moves round the Sun on an ellipse with the Sun at one focus, and every satellite round the Earth with the Earth's centre at one focus. In hospitals, a lithotripter breaks kidney stones without surgery: a shock wave is produced at one focus of an elliptical reflector and every part of it bounces off the reflector to meet at the other focus, where the stone is placed. In this section we turn the string rule into the equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, meet the number $e$ — the eccentricity — that measures how stretched an ellipse is, and learn to find the equation of an ellipse from given data.
Two Foci and a Constant Sum
An ellipse is the set of all points in a plane the sum of whose distances from two fixed points $F_1$ and $F_2$ of the plane is a constant, greater than the distance $F_1F_2$. The fixed points are the foci (singular: focus). The mid-point of $F_1F_2$ is the centre of the ellipse. The chord through the foci is the major axis, and its two ends are the vertices of the ellipse. The chord through the centre perpendicular to the major axis is the minor axis.
We write the lengths as follows: major axis $2a$, minor axis $2b$, distance between the foci $2c$. So $a$ is the distance from the centre to a vertex, $b$ the distance from the centre to an end of the minor axis, and $c$ the distance from the centre to a focus. Why must the constant exceed $F_1F_2$? If it equalled $F_1F_2$, the only points would be those on the segment $F_1F_2$; if it were smaller, there would be no points at all, because $PF_1+PF_2\ge F_1F_2$ for every point $P$ (the triangle inequality).
The constant sum is $2a$. Let $A$ be the vertex beyond $F_2$. Then $AF_2=a-c$ and $AF_1=a+c$, so $AF_1+AF_2=2a$. Since the sum is the same for every point, every point $P$ of the ellipse satisfies $$PF_1+PF_2=2a.$$
For every ellipse, $$a^2=b^2+c^2,\qquad\text{i.e.}\quad c^2=a^2-b^2 .$$ In particular $a\gt b$ and $a\gt c$: the major axis really is the longer one.
Proof. Let $B$ be an end of the minor axis and $O$ the centre. $B$ is on the perpendicular bisector of $F_1F_2$, so $BF_1=BF_2$; since $BF_1+BF_2=2a$, each equals $a$. In the right triangle $OBF_2$ the legs are $OB=b$ and $OF_2=c$ and the hypotenuse is $BF_2=a$. By Pythagoras, $a^2=b^2+c^2$. $\blacksquare$
So the three numbers form a right triangle: half the minor axis and the focal distance $c$ are the legs, and the semi-major axis $a$ is the hypotenuse. Knowing any two of $a$, $b$, $c$ gives the third.
The eccentricity of an ellipse is the ratio $$e=\frac ca=\frac{\sqrt{a^2-b^2}}{a}.$$ Since $0\lt c\lt a$, every ellipse has $0\lt e\lt1$.
The eccentricity measures how far the foci are from the centre, as a fraction of $a$. If the foci move together to the centre ($c\to0$) then $b\to a$ and the ellipse becomes a circle — so a circle can be thought of as an ellipse with $e=0$. If the foci move out towards the vertices ($c\to a$) then $b\to0$ and the ellipse flattens towards the segment $F_1F_2$. These two limiting cases are only remarks; every ellipse we work with has $0\lt e\lt1$.
(b) An ellipse has semi-minor axis $b=5$ and eccentricity $\tfrac{12}{13}$. Find $a$ and $c$.
The Standard Equation of an Ellipse
Choose the axes so that the centre is the origin and the foci lie on the $x$-axis: $F_1(-c,0)$ and $F_2(c,0)$. The vertices are then $(\pm a,0)$ and the ends of the minor axis $(0,\pm b)$.
The ellipse with foci $(\pm c,0)$ and constant sum $2a$ ($a\gt c\ge0$) has equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad b^2=a^2-c^2.$$
Proof. $P(x,y)$ is on the ellipse exactly when $PF_1+PF_2=2a$: $$\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a.$$ Move the second root across and square: $$\begin{aligned}(x+c)^2+y^2&=4a^2-4a\sqrt{(x-c)^2+y^2}\\&\quad+(x-c)^2+y^2.\end{aligned}$$ The $x^2$, $y^2$, $c^2$ cancel, leaving $4cx-4a^2=-4a\sqrt{(x-c)^2+y^2}$, i.e. $$a\sqrt{(x-c)^2+y^2}=a^2-cx.$$ Square again: $a^2(x^2-2cx+c^2+y^2)=a^4-2a^2cx+c^2x^2$, so $(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$. Put $b^2=a^2-c^2$ and divide by $a^2b^2$: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.
Conversely, if $(x,y)$ satisfies the equation then $|x|\le a$ and $y^2=b^2\left(1-\tfrac{x^2}{a^2}\right)$. Substituting, and using $b^2+c^2=a^2$, $$\begin{aligned}PF_2^2&=(x-c)^2+b^2-\tfrac{b^2}{a^2}x^2\\&=\tfrac{c^2}{a^2}x^2-2cx+a^2=\left(a-\tfrac cax\right)^2.\end{aligned}$$ As $|x|\le a$ and $c\lt a$, $a-\tfrac cax\gt0$, so $PF_2=a-\tfrac cax$. In the same way $PF_1=a+\tfrac cax$, and $PF_1+PF_2=2a$. So the equation describes exactly the ellipse. $\blacksquare$
If instead the foci are on the $y$-axis at $(0,\pm c)$, the roles of $x$ and $y$ swap and the equation is $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$. In both forms $a\gt b$, so here is the rule for reading an equation: the larger denominator belongs to the major axis. If the larger number is under $x^2$, the major axis is along the $x$-axis; if it is under $y^2$, the major axis is along the $y$-axis.
| Equation ($a\gt b\gt0$) | Major axis | Vertices | Ends of minor axis | Foci |
|---|---|---|---|---|
| $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ | along $x$-axis, length $2a$ | $(\pm a,0)$ | $(0,\pm b)$ | $(\pm c,0)$ |
| $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ | along $y$-axis, length $2a$ | $(0,\pm a)$ | $(\pm b,0)$ | $(0,\pm c)$ |
In both cases $c^2=a^2-b^2$, $e=\dfrac ca$, and the ellipse is symmetric about both axes and about the centre: if $(x,y)$ is on it, so are $(-x,y)$, $(x,-y)$ and $(-x,-y)$, because only $x^2$ and $y^2$ appear.
A latus rectum of an ellipse is a chord through a focus perpendicular to the major axis. For $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ the latus rectum through $(c,0)$ has ends $\left(c,\pm\dfrac{b^2}{a}\right)$, so its length is $$\frac{2b^2}{a}.$$ (There are two latera recta, one through each focus, of the same length.)
Proof. Put $x=c$: $\dfrac{y^2}{b^2}=1-\dfrac{c^2}{a^2}=\dfrac{a^2-c^2}{a^2}=\dfrac{b^2}{a^2}$, so $y^2=\dfrac{b^4}{a^2}$ and $y=\pm\dfrac{b^2}{a}$. The length is $2\cdot\dfrac{b^2}{a}$. $\blacksquare$
(a) vertices $(\pm13,0)$ and foci $(\pm5,0)$; (b) foci $(0,\pm3)$ and major axis of length $10$; (c) foci $(\pm4,0)$ and eccentricity $\tfrac23$.
Ellipses at Work: Orbits and Arches
A satellite moves on an ellipse with the centre of the Earth at one focus $F$. Its closest point to the Earth's centre, the perigee, is the vertex nearer $F$, at distance $a-c$; its farthest point, the apogee, is the other vertex, at distance $a+c$. (For orbits round the Sun the words are perihelion and aphelion.) Adding and subtracting, $$a=\frac{r_{\min}+r_{\max}}{2},\qquad c=\frac{r_{\max}-r_{\min}}{2},\qquad e=\frac{r_{\max}-r_{\min}}{r_{\max}+r_{\min}},$$ and $b^2=a^2-c^2=(a-c)(a+c)=r_{\min}\,r_{\max}$. The number $a$ is called the semi-major axis of the orbit.
1. Using $c^2=a^2+b^2$. For an ellipse the semi-major axis $a$ is the hypotenuse: $c^2=a^2-b^2$. (The plus sign belongs to the hyperbola in Section 10.4.)
2. Assuming $a$ is under $x^2$. In $\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$, $a^2=25$ and the major axis is along the $y$-axis; the foci are $(0,\pm3)$, not $(\pm3,0)$.
3. Forgetting to make the right side $1$. $4x^2+9y^2=36$ must be divided by $36$ to give $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ before reading $a$ and $b$.
4. Mixing up lengths and half-lengths. "Major axis $10$" means $a=5$; "foci $6$ apart" means $c=3$. The latus rectum is $\dfrac{2b^2}{a}$, using the half-lengths.
5. Measuring orbits from the surface. In orbit problems, $a\pm c$ are distances from the Earth's centre; add the Earth's radius to heights above the surface.
In JEE Advanced 2026 (foci, directrices and eccentricity of conics); not in JEE Main 2026 or the CBSE year-end exam. JEE Advanced often uses the focal distances of an ellipse and its directrices.
Focal distances. The proof of Theorem 10.6 showed that for $P(x,y)$ on $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $$PF_1=a+ex,\qquad PF_2=a-ex,$$ where $F_1=(-c,0)$, $F_2=(c,0)$. They add to $2a$, and $PF_1\cdot PF_2=a^2-e^2x^2$.
Directrices. Write $PF_2=a-ex=e\left(\tfrac ae-x\right)$. The bracket is the distance of $P$ from the line $x=\tfrac ae$. So every point of the ellipse satisfies $PF_2=e\cdot PM_2$, where $PM_2$ is its distance from the directrix $x=\tfrac ae$ (and similarly with $F_1$ and $x=-\tfrac ae$) — the focus–directrix picture of the chapter teaser, with $e\lt1$.
Worked example. For $P\left(4,\tfrac{12}{5}\right)$ on $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, find $PF_1$ and $PF_2$ and check $PF_2=e\cdot PM_2$. Solution. $a=5$, $c=3$, $e=\tfrac35$: $PF_1=5+\tfrac{12}{5}=\tfrac{37}{5}$, $PF_2=5-\tfrac{12}{5}=\tfrac{13}{5}$. Directly: $PF_2=\sqrt{1+\tfrac{144}{25}}=\tfrac{13}{5}$ ✓. The directrix is $x=\tfrac{25}{3}$, so $PM_2=\tfrac{25}{3}-4=\tfrac{13}{3}$ and $e\cdot PM_2=\tfrac35\cdot\tfrac{13}{3}=\tfrac{13}{5}$ ✓.
Exercises
- 1 $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$
- 2 $\dfrac{x^2}{4}+\dfrac{y^2}{25}=1$
- 3 $\dfrac{x^2}{100}+\dfrac{y^2}{64}=1$
- 4 $25x^2+16y^2=400$
- 5 $x^2+4y^2=4$
- 6 $9x^2+5y^2=45$
- 7 $3x^2+2y^2=18$
- 8 Find the ends of both latera recta of $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$.
- 1 Vertices $(\pm7,0)$, foci $(\pm5,0)$.
- 2 Vertices $(0,\pm10)$, foci $(0,\pm8)$.
- 3 Ends of the major axis $(\pm6,0)$, ends of the minor axis $(0,\pm2)$.
- 4 Major axis of length $26$, foci $(\pm12,0)$.
- 5 Foci $(0,\pm4)$, minor axis of length $6$.
- 6 Foci $(\pm2,0)$, eccentricity $\tfrac12$.
- 7 Vertices $(0,\pm6)$, eccentricity $\tfrac13$.
- 8 Major axis along the $x$-axis, $b=3$, passing through $\left(4,\tfrac95\right)$.
- 9 Major axis along the $y$-axis, passing through $(3,2)$ and $(1,6)$.
- 10 Foci $(\pm3,0)$, passing through $(4,1)$.
- 1 The minor axis of an ellipse is half its major axis. Find its eccentricity.
- 2 The latus rectum of an ellipse is one third of its major axis. Find its eccentricity.
- 3 The distance between the foci of an ellipse equals the length of its minor axis. Find its eccentricity.
- 4 Show that $P\left(3,\tfrac{16}{5}\right)$ lies on $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, and verify that $PF_1+PF_2=10$.
- 5 A point moves so that the sum of its distances from $(4,0)$ and $(-4,0)$ is always $10$. Find the equation of its path.
- 6 A semi-elliptical arch over a canal is $20\ \text{m}$ wide and $6\ \text{m}$ high at the centre. Find its height $2\ \text{m}$ from one end.
- 7 The Earth's orbit has semi-major axis about $149.6$ million km and eccentricity about $0.0167$. Find the Earth's least and greatest distances from the Sun, and the distance between the foci.
- 8 A lithotripter's reflector is part of an ellipse with $a=15\ \text{cm}$ and $b=9\ \text{cm}$. The shock wave starts at one focus. How far from the source must the kidney stone be?
- 9 Find the ellipse, centred at the origin with major axis along the $x$-axis, whose latus rectum is $5$ and eccentricity is $\tfrac12$.
How this is examined: Board papers ask for the eccentricity or foci of a given ellipse (1 mark), its equation from vertices, foci or eccentricity (2 marks), and the full list — foci, vertices, axes, eccentricity, latus rectum — for an equation such as $36x^2+4y^2=144$ (3 marks); case studies use arches and orbits. JEE links $e$ to the latus rectum and axes and uses the focal distances $a\pm ex$.
E1. Board · 1 mark The eccentricity of $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ is (A) $\tfrac35$ (B) $\tfrac45$ (C) $\tfrac53$ (D) $\tfrac34$
E2. Board · 2 marks Find the equation of the ellipse with vertices $(0,\pm5)$ and foci $(0,\pm4)$.
E3. Board · 3 marks Find the foci, the vertices, the lengths of the major and minor axes, the eccentricity and the length of the latus rectum of $36x^2+4y^2=144$.
E4. Board · Case study A road tunnel has a semi-elliptical entrance $16\ \text{m}$ wide at road level and $4\ \text{m}$ high at the centre. Take the road as the $x$-axis and the centre of the base as the origin. (i) Write the equation of the ellipse. (1 mark) (ii) Find the height of the tunnel $4\ \text{m}$ from the centre line. (1 mark) (iii) A truck $6\ \text{m}$ wide drives exactly along the centre line. Find the greatest height it can have to pass through. (2 marks)
E5. JEE Main The latus rectum of an ellipse is equal to half its minor axis. Its eccentricity is (A) $\tfrac12$ (B) $\tfrac{\sqrt3}{2}$ (C) $\tfrac{1}{\sqrt2}$ (D) $\tfrac34$
E6. JEE Advanced Numerical answer. $F_1$ and $F_2$ are the foci of $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ and $P$ moves on the ellipse. Find the greatest value of $PF_1\cdot PF_2$.
Research question. How do the string length and the distance between the pins control the shape of an ellipse — and what do real planetary and comet orbits look like when drawn to scale?
Why it matters. Astronomers describe an orbit by just two numbers, the semi-major axis $a$ and the eccentricity $e$. The Earth's orbit ($e\approx0.017$) looks like a perfect circle on paper, Mars's ($e\approx0.093$) almost so, while Halley's comet ($e\approx0.967$) swings from inside Venus's orbit to beyond Neptune's. Drawing them yourself shows why Kepler needed very precise data to notice that orbits are not circles.
What to do.
- On a sheet of card, push in two pins $2c$ apart and tie a loop of string with total length $2a+2c$ (so that the part going round the pencil is $2a$). Draw the ellipse with a pencil keeping the string tight.
- Measure the major and minor axes of your drawing. Check that $b^2=a^2-c^2$ and compute $e=\tfrac ca$.
- Repeat with the same string and pins $0$, $2$, $4$, $6$ and $8\ \text{cm}$ apart. Plot $\tfrac ba$ against $e$ and compare with the curve $\tfrac ba=\sqrt{1-e^2}$.
- Choose a scale (say $a=8\ \text{cm}$) and draw the orbits of the Earth, Mars and Halley's comet using their eccentricities, marking the Sun at a focus.
Tools: Card, two pins, string, a pencil and ruler (or a GeoGebra-style app with its ellipse tool), a spreadsheet
Hand in: Your drawings, a table of $a$, $b$, $c$, $e$ for each, the graph of $\tfrac ba$ against $e$, and your three scale orbits with a sentence on which ones look like circles and why.
Going further. For the Earth, compute the perihelion and aphelion distances $a(1\mp e)$ and the ratio of sunlight intensity at the two (intensity varies as $\tfrac1{r^2}$). Is the Earth's distance from the Sun the reason for the seasons?
🧪 Self-check quiz
Q1. The major axis of $\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ lies along
Q2. For $\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$, the distance from the centre to each focus is
Q3. The eccentricity $e$ of an ellipse always satisfies
Q4. The length of the latus rectum of $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ is
Q5. For every point $P$ of $\dfrac{x^2}{36}+\dfrac{y^2}{20}=1$, the sum of its distances from the two foci is