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Vectors and conics

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Attempts to answer problems in areas as diverse as science, technology and economics involve solving simultaneous linear equations. In this free course, Vectors and conics, we look at some of the equations that represent points, lines and planes in mathematics. We explore concepts such as Euclidean space, vectors, dot products and conics.

After studying this course, you should be able to:

  • recognise the equation of a line in the plane
  • determine the point of intersection of two lines in the plane, if it exists
  • recognise the one-one correspondence between the set of points in three-dimensional space and the set of ordered triples of real numbers
  • recognise the equation of a plane in three dimensions
  • explain what are meant by a vector, a scalar multiple of a vector, and the sum and difference of two vectors.

By: The Open University

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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 .

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