10.4 The Hyperbola
A hyperbola is the set of points whose distances from two foci differ by a constant $2a$. With the foci at $(\pm c,0)$ it is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, where now $c^2=a^2+b^2$ and the eccentricity $e=\dfrac ca$ is greater than $1$.
Two radio transmitters on a coast, $100\ \text{km}$ apart, send out a pulse at exactly the same instant. A ship at sea picks up the pulse from the nearer transmitter a fraction of a second before the pulse from the other one. Radio waves travel at a known speed, so the delay tells the ship by how much its distance from one transmitter exceeds its distance from the other. That does not fix the ship's position — but it does put the ship on a definite curve, the set of all points whose distances from the two transmitters differ by that amount. That curve is a hyperbola, and a second pair of transmitters puts the ship on a second hyperbola; where the two cross, the ship is. This was the idea behind radio navigation systems used for decades before satellites, and the same "time difference of arrival" method is used today to locate earthquakes, lightning strikes and mobile phones.
The ellipse kept the sum of the two focal distances fixed; the hyperbola keeps their difference fixed. That one change of sign runs through this whole section. The curve now has two separate pieces, its branches; the equation becomes $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$; the relation between $a$, $b$, $c$ becomes $c^2=a^2+b^2$; and the eccentricity, which was less than $1$ for an ellipse, is now greater than $1$. Hyperbolas also appear in the shape of power-station cooling towers and in the paths of comets and interstellar objects that pass the Sun once and never return.
A Constant Difference of Distances
A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points $F_1$ and $F_2$ of the plane is a constant: $|PF_1-PF_2|$ is the same for every point $P$, and this constant is positive and less than $F_1F_2$. The fixed points are the foci; the mid-point of $F_1F_2$ is the centre. The line through the foci meets the hyperbola at two points, its vertices; the segment joining the vertices is the transverse axis, and the line through the centre perpendicular to it is the direction of the conjugate axis.
The points nearer $F_2$ (where $PF_1-PF_2$ is the positive constant) form one branch of the hyperbola; the points nearer $F_1$ form the other. As for the ellipse we write $2c$ for the distance between the foci and $2a$ for the length of the transverse axis, so the vertices are $a$ from the centre and the foci $c$ from it, with now $c\gt a$. At the vertex $A$ near $F_2$: $AF_1-AF_2=(c+a)-(c-a)=2a$. So the constant difference is $2a$: $$|PF_1-PF_2|=2a .$$ Why must the constant be less than $F_1F_2=2c$? In any triangle $PF_1F_2$ one side is longer than the difference of the other two, so $|PF_1-PF_2|\lt F_1F_2$ for points off the line $F_1F_2$. A constant equal to $2c$ would give only two rays of that line, and a larger one no points at all.
Since $c\gt a$, we can define a positive number $b$ by $$b^2=c^2-a^2,\qquad\text{i.e.}\quad c^2=a^2+b^2 .$$ The conjugate axis is the segment of length $2b$ on that perpendicular line, centred at the centre. Unlike the minor axis of an ellipse it does not meet the curve, but $b$ is just as important: it shapes the equation and controls how widely the branches open.
The eccentricity of a hyperbola is $$e=\frac ca=\frac{\sqrt{a^2+b^2}}{a}.$$ Since $c\gt a$, every hyperbola has $e\gt1$.
For a fixed $a$, a larger eccentricity means larger $b$ and branches that open more widely; an eccentricity close to $1$ gives narrow branches hugging the transverse axis. When $a=b$ the hyperbola is called a rectangular hyperbola (or equilateral hyperbola); then $c^2=2a^2$ and $e=\sqrt2$.
(b) A hyperbola has $a=3$ and eccentricity $2$. Find $c$ and $b$.
The Standard Equation of a Hyperbola
Put the centre at the origin and the foci on the $x$-axis: $F_1(-c,0)$, $F_2(c,0)$. The vertices are then $(\pm a,0)$.
The hyperbola with foci $(\pm c,0)$ and constant difference $2a$ ($0\lt a\lt c$) has equation $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad b^2=c^2-a^2.$$
Proof. $P(x,y)$ is on the hyperbola exactly when $PF_1-PF_2=\pm2a$: $$\sqrt{(x+c)^2+y^2}=\pm2a+\sqrt{(x-c)^2+y^2}.$$ Squaring, $$\begin{aligned}(x+c)^2+y^2&=4a^2\pm4a\sqrt{(x-c)^2+y^2}\\&\quad+(x-c)^2+y^2,\end{aligned}$$ and after cancelling, $4cx-4a^2=\pm4a\sqrt{(x-c)^2+y^2}$, i.e. $cx-a^2=\pm a\sqrt{(x-c)^2+y^2}$. Squaring again: $c^2x^2-2a^2cx+a^4=a^2(x^2-2cx+c^2+y^2)$, so $(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$. With $b^2=c^2-a^2$, dividing by $a^2b^2$ gives $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$.
Conversely, if $(x,y)$ satisfies the equation then $x^2\ge a^2$ and $y^2=b^2\left(\tfrac{x^2}{a^2}-1\right)$, so, using $a^2+b^2=c^2$, $$\begin{aligned}PF_2^2&=(x-c)^2+\tfrac{b^2}{a^2}x^2-b^2\\&=\tfrac{c^2}{a^2}x^2-2cx+a^2=\left(\tfrac cax-a\right)^2,\end{aligned}$$ and similarly $PF_1^2=\left(\tfrac cax+a\right)^2$. For $x\ge a$ (so $\tfrac cax\gt a$) this gives $PF_1-PF_2=\left(\tfrac cax+a\right)-\left(\tfrac cax-a\right)=2a$; for $x\le-a$ it gives $PF_2-PF_1=2a$. So the equation describes exactly the hyperbola. $\blacksquare$
If the foci are on the $y$-axis at $(0,\pm c)$, the equation is $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$. Notice how the rule for reading the equation differs from the ellipse: the variable with the positive term names the transverse axis, and $a^2$ is always the denominator under the positive term — even if it is the smaller number. In a hyperbola $a$ may be smaller than, equal to, or larger than $b$.
| Equation ($a,b\gt0$) | Transverse axis | Vertices | Foci | Branches open |
|---|---|---|---|---|
| $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ | along $x$-axis, length $2a$ | $(\pm a,0)$ | $(\pm c,0)$ | left and right |
| $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ | along $y$-axis, length $2a$ | $(0,\pm a)$ | $(0,\pm c)$ | up and down |
In both cases $c^2=a^2+b^2$, $e=\dfrac ca\gt1$, the conjugate axis has length $2b$, and the curve is symmetric about both axes and the centre. The lines $y=\pm\dfrac bax$ (for the first form), the diagonals of the rectangle with corners $(\pm a,\pm b)$, are the asymptotes of the hyperbola: far from the centre the branches come closer and closer to them without meeting them. They are not part of the Board syllabus as a topic, but drawing the rectangle and its diagonals first is the quickest way to sketch a hyperbola.
The chord of $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ through a focus perpendicular to the transverse axis (a latus rectum) has ends $\left(\pm c,\pm\dfrac{b^2}{a}\right)$ and length $$\frac{2b^2}{a}.$$
Proof. Put $x=c$: $\dfrac{y^2}{b^2}=\dfrac{c^2}{a^2}-1=\dfrac{c^2-a^2}{a^2}=\dfrac{b^2}{a^2}$, so $y=\pm\dfrac{b^2}{a}$ and the length is $\dfrac{2b^2}{a}$; by symmetry the same holds at $x=-c$. $\blacksquare$
(a) vertices $(\pm4,0)$ and foci $(\pm6,0)$; (b) foci $(0,\pm5)$ and latus rectum of length $\tfrac92$; (c) foci $(\pm4,0)$ and eccentricity $2$.
Hyperbolas at Work and Telling the Conics Apart
(a) Find the equation of the curve on which the ship lies, taking the mid-point of $AB$ as the origin and $AB$ along the $x$-axis with $B$ on the positive side.
(b) The ship is also known to be $40\ \text{km}$ out to sea from the coast. Find its position.
By now you have met all four standard conics. When an equation is given, put it in standard form (constant term moved to the right and made equal to $1$, or a single squared term) and look at its shape:
1. Only one variable squared ($y^2=4ax$, $x^2=-4ay$, …): a parabola. The non-squared variable names the axis; the sign gives the side.
2. Both squared, same sign, equal coefficients ($x^2+y^2=r^2$): a circle.
3. Both squared, same sign, different coefficients ($\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$): an ellipse; the larger denominator marks the major axis, $c^2=a^2-b^2$, $e\lt1$.
4. Both squared, opposite signs ($\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1$): a hyperbola; the positive term marks the transverse axis, $c^2=a^2+b^2$, $e\gt1$.
1. Using the ellipse relation. For a hyperbola $c^2=a^2+b^2$: the focal distance $c$ is the largest of the three. $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ has $c=5$, not $\sqrt7$.
2. Taking $a$ as the larger number. In $\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$, $a^2=9$ (under the positive term), not $16$; the transverse axis is along the $y$-axis.
3. Forgetting the absolute value. The rule is $|PF_1-PF_2|=2a$. Without the absolute value you get only one branch.
4. An eccentricity below $1$. If your hyperbola has $e\lt1$, you have used $c^2=a^2-b^2$ somewhere.
5. Mixing up the constants. The difference of the focal distances is $2a$ (the transverse axis), not $2c$ (the distance between the foci) and not $2b$.
In JEE Advanced 2026 (equations of tangents to conics); not in JEE Main 2026 or the CBSE year-end exam. JEE Advanced asks when a line $y=mx+c$ touches $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. (Here $c$ is the intercept of the line, not the focal distance.)
Method. Substitute and clear denominators: $(b^2-a^2m^2)x^2-2a^2mc\,x-a^2(c^2+b^2)=0$. If $m\ne\pm\tfrac ba$ this is a quadratic, and the line touches the hyperbola when its discriminant is $0$: $4a^4m^2c^2+4a^2(b^2-a^2m^2)(c^2+b^2)=0$. Dividing by $4a^2$ and expanding, the $a^2m^2c^2$ terms cancel and we are left with $b^2(c^2+b^2-a^2m^2)=0$: $$y=mx+c\ \text{touches}\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\iff c^2=a^2m^2-b^2 .$$ (So no tangent has $|m|\lt\tfrac ba$: compare $c^2=a^2m^2+b^2$ for the ellipse.)
Worked example. Find $c$ so that $y=2x+c$ touches $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$. Solution. $c^2=9\cdot4-4=32$, $c=\pm4\sqrt2$. Check ($c=4\sqrt2$): the quadratic becomes $-32x^2-144\sqrt2\,x-324=0$, i.e. $8x^2+36\sqrt2\,x+81=0$, with discriminant $2592-2592=0$ ✓; the point of contact is $\left(-\tfrac{9\sqrt2}{4},-\tfrac{\sqrt2}{2}\right)$.
Exercises
- 1 $\dfrac{x^2}{25}-\dfrac{y^2}{144}=1$
- 2 $\dfrac{y^2}{16}-\dfrac{x^2}{48}=1$
- 3 $4x^2-y^2=16$
- 4 $25y^2-9x^2=225$
- 5 $x^2-y^2=9$
- 6 $3x^2-4y^2=12$
- 7 $9y^2-x^2=9$
- 8 Find the ends of both latera recta of $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$.
- 1 Vertices $(\pm5,0)$, foci $(\pm7,0)$.
- 2 Vertices $(0,\pm6)$, foci $(0,\pm10)$.
- 3 Foci $(\pm8,0)$, transverse axis of length $10$.
- 4 Foci $(0,\pm13)$, conjugate axis of length $24$.
- 5 Foci $(\pm\sqrt{13},0)$, latus rectum of length $9$.
- 6 Foci $(\pm9,0)$, eccentricity $3$.
- 7 Vertices $(\pm4,0)$, passing through $(8,3\sqrt3)$.
- 8 Vertices $(\pm5,0)$, eccentricity $\sqrt2$.
- 9 Transverse axis along the $x$-axis, passing through $(5,0)$ and $(10,3\sqrt3)$.
- 1 The latus rectum of a hyperbola is half its transverse axis. Find its eccentricity.
- 2 The conjugate axis of a hyperbola equals its transverse axis. Find its eccentricity.
- 3 The distance between the foci of a hyperbola is three times the distance between its vertices. Find its eccentricity, and the ratio $\tfrac ba$.
- 4 Show that $P\left(5,\tfrac{16}{3}\right)$ lies on $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ and verify that $|PF_1-PF_2|=6$.
- 5 A point moves so that the difference of its distances from $(0,5)$ and $(0,-5)$ is always $8$. Find the equation of its path.
- 6 Two listening posts $120\ \text{km}$ apart hear an explosion; the sound reaches post $B$ before post $A$, showing that the explosion is $72\ \text{km}$ nearer $B$. Taking the mid-point as origin and $B$ at $(60,0)$, find the curve on which the explosion lies.
- 7 The waist (narrowest part) of a cooling tower is $40\ \text{m}$ across. Its vertical cross-section is a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with the $y$-axis vertical through the centre of the waist. At $30\ \text{m}$ above the waist the tower is $50\ \text{m}$ across. Find the equation of the cross-section.
- 8 Show that, for $k\lt9$, $\dfrac{x^2}{16-k}+\dfrac{y^2}{9-k}=1$ is an ellipse and for $9\lt k\lt16$ it is a hyperbola, and that in every case the foci are $(\pm\sqrt7,0)$.
- 9 Name each curve: (a) $3x^2+3y^2=12$; (b) $3x^2-3y^2=12$; (c) $3x^2+5y^2=15$; (d) $3y^2=12x$.
How this is examined: Board papers ask for the eccentricity or foci of a given hyperbola (1 mark), its equation from vertices and foci (2 marks), and the full list of features for an equation such as $64x^2-36y^2=2304$, or an equation from the foci and latus rectum (3 marks). JEE mixes ellipses and hyperbolas with the same foci and uses the tangency condition.
E1. Board · 1 mark The eccentricity of $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ is (A) $\tfrac43$ (B) $\tfrac53$ (C) $\tfrac54$ (D) $\tfrac{\sqrt7}{3}$
E2. Board · 2 marks Find the equation of the hyperbola with vertices $(0,\pm5)$ and foci $(0,\pm8)$.
E3. Board · 3 marks Find the foci, the vertices, the lengths of the transverse and conjugate axes, the eccentricity and the length of the latus rectum of $64x^2-36y^2=2304$.
E4. Board · Case study The cooling tower of a thermal power station has a hyperbolic vertical cross-section. Its waist (narrowest part) is $60\ \text{m}$ across, and $40\ \text{m}$ above the waist the tower is $100\ \text{m}$ across. Take the centre of the waist as the origin and the axis of the tower as the $y$-axis, so the cross-section is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. (i) Find $a$. (1 mark) (ii) Find $b$ and the equation. (1 mark) (iii) The base of the tower is $60\ \text{m}$ below the waist. Find the diameter of the base. (2 marks)
E5. JEE Main The ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ and the hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{b^2}=1$ are confocal (they have the same foci). Then $b^2$ equals (A) $5$ (B) $21$ (C) $9$ (D) $12$
E6. JEE Advanced One or more options correct. For the hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$, which statements are true? (A) its eccentricity is $2$ (B) its latus rectum has length $12$ (C) its foci are $(\pm4,0)$ (D) the line $y=x+2\sqrt2$ touches it
Research question. Can two phones locate a hand-clap — and how accurately does the "difference of distances" put the source on a hyperbola?
Why it matters. Seismologists locate earthquakes, meteorologists locate lightning and engineers locate leaks in water pipes from the differences in arrival times of a signal at several sensors. Each time difference puts the source on one branch of a hyperbola with two sensors as foci; two or more hyperbolas cross at the source. You can test the method with sound, which travels at about $343\ \text{m/s}$ in air at $20^\circ\text{C}$.
What to do.
- In a playground, place two phones $A$ and $B$ about $20\ \text{m}$ apart, both recording sound (use an audio app that shows the waveform). Mark the line $AB$ with chalk and choose axes with the origin at its mid-point.
- A friend stands at a measured point $P$ (not on $AB$) and claps sharply once. On each recording, find the time of the clap relative to a common start signal (for example a whistle blown exactly midway between the phones).
- Compute $\Delta t$ and $2a=343\,\Delta t$, then the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with $c=10\ \text{m}$. Does $P$ lie on the correct branch? Compare the measured $|PA-PB|$ with $343\,\Delta t$.
- Repeat from three or four different points, and from one point on the perpendicular bisector of $AB$ (what should $\Delta t$ be there?).
Tools: Two phones with an audio-recording app, a measuring tape, chalk, a whistle, a spreadsheet
Hand in: A sketch of the set-up with your hyperbolas drawn, a table of predicted and measured $|PA-PB|$, and a paragraph explaining the main sources of error and how a third phone would pin down the exact position.
Going further. With a third phone $C$, each pair of phones gives a hyperbola. Find the intersection of two of them numerically (a spreadsheet or graphing app helps) and compare it with the true position of the clapper.
🧪 Self-check quiz
Q1. For a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with foci $(\pm c,0)$,
Q2. The eccentricity of the rectangular hyperbola $x^2-y^2=1$ is
Q3. The transverse axis of $\dfrac{y^2}{4}-\dfrac{x^2}{25}=1$ lies along
Q4. For every point $P$ of $\dfrac{x^2}{49}-\dfrac{y^2}{24}=1$, $|PF_1-PF_2|$ equals
Q5. Which equation represents a hyperbola?