Vectors and conics
Vectors and conics

This free course is available to start right now. Review the full course description and key learning outcomes and create an account and enrol if you want a free statement of participation.

Free course

Vectors and conics

4.6 Hyperbola (e > 1)

A hyperbola is the set of points P in the plane whose distances from a fixed point F are e times their distances from a fixed line d, where e > 1. We obtain a hyperbola in standard form if

  1. the focus F lies on the x-axis, and has coordinates (ae, 0), where a > 0;

  2. the directrix d is the line with equation x = a/e.

Let P (x, y) be an arbitrary point on the hyperbola, and let M be the foot of the perpendicular from P to the directrix. Since FP = e × PM, by the definition of the hyperbola, it follows that FP2 = e2 × PM2; that is,

Multiplying out the brackets, we obtain

x2 − 2aex + a2e2 + y2 = e2x2 − 2aex + a2,

which simplifies to the equation

x2(e2 − 1) − y2 = a2(e2 − 1),

that is,

Substituting b for , so that b2 = a2(e2 − 1), we obtain the standard form of the equation of the hyperbola

This equation is symmetric in x and in y, so that the hyperbola also has a second focus F′ at (−ae, 0), and a second directrix d′ with equation x = −a/e.

The hyperbola intersects the x-axis at the points (±a, 0). We call the line segment joining the points (±a, 0) the major axis or transverse axis of the hyperbola, and the line segment joining the points (0, ±b) the minor axis or conjugate axis of the hyperbola (this is not a chord of the hyperbola). The origin is the centre of this hyperbola.

Each point with coordinates (a sec t, b tan t) lies on the hyperbola, since

Note: In general, sec2t = 1 + tan2t.

Then, just as for the parabola, we can check that

gives a parametric representation of the hyperbola.

Note: An alternative parametrisation, using hyperbolic functions, is x = acosht, y = bsinht (t).

Two other features of the shape of the hyperbola stand out.

First, the hyperbola consists of two separate curves or branches.

Secondly, the lines with equations y = ±bx/a divide the plane into two pairs of opposite sectors; the branches of the hyperbola lie in one pair. As x → ±∞, the branches of the hyperbola get closer and closer to these two lines. We call the lines y = ±bx/a the asymptotes of the hyperbola.

We summarise these facts as follows:

Hyperbola in standard form

A hyperbola in standard form has equation

It can also be described by the parametric equations

It has foci (±ae, 0) and directrices x = ±a/e; its major axis is the line segment joining the points (±a, 0), and its minor axis is the line segment joining the points (0, ±b).

Example 57

Let P be a point , t, on the hyperbola with equation x2 − 2y2 = 1.

  • (a) Determine the foci F and F′ of the hyperbola.

  • (b) Determine the gradients of FP and FP, when these lines are not parallel to the y-axis.

  • (c) Find the point P on the hyperbola, in the first quadrant, for which FP is perpendicular to FP.


  • (a) This hyperbola is of the form with a = 1 and , so . If e denotes the eccentricity of the hyperbola, so that b2 = a2(e2 − 1), we have

  • it follows that , so

  • In the general case, the foci are (±ae, 0); it follows that here the foci are .

  • (b) Let F and F′ be and , respectively. (It does not matter which way round these are chosen.)

  • Then the gradient of FP is

  • where we know that , since FP is not parallel to the y-axis.

  • Similarly, the gradient of FP is

  • where we know that , since FP is not parallel to the y-axis.

  • (c) When FP is perpendicular to FP, we have

  • We may rewrite this in the form

  • so 2sec2t − 3 + tan2t = 0.

  • Since sec2t = 1 + tan2t, it follows that 3 tan2t = 1.

  • Since we are looking for a point P in the first quadrant, we choose .

  • When , we have . Since we are looking for a point P in the first quadrant, we choose .

  • It follows that the required point P has coordinates .


Take your learning further

Making the decision to study can be a big step, which is why you'll want a trusted University. The Open University has 50 years’ experience delivering flexible learning and 170,000 students are studying with us right now. Take a look at all Open University courses.

If you are new to university level study, find out more about the types of qualifications we offer, including our entry level Access courses and Certificates.

Not ready for University study then browse over 900 free courses on OpenLearn and sign up to our newsletter to hear about new free courses as they are released.

Every year, thousands of students decide to study with The Open University. With over 120 qualifications, we’ve got the right course for you.

Request an Open University prospectus