from The Open University
Alternatively you can skip the navigation by pressing 'Enter'.

TV, Radio & Events menu item
Coming up
Thinking Allowed: The Ethnography Award winner 2015
Monday, 27th April 2015 00:15  BBC FourIn this episode of BBC Radio 4's Thinking Allowed, Laurie Taylor announces the Ethnography award winner 2015.... Read more: Thinking Allowed: The Ethnography Award winner 2015Christopher Plummer on King Lear
Tuesday, 28th April 2015 04:35  Sky Arts 1Secret History Of... Camberwell Gve
Tuesday, 28th April 2015 20:00  BBC FourWastemen: The Home Front
Tuesday, 28th April 2015 21:00  BBC TwoCatch up
Secret History Of... Deptford High St
Available until Friday, 22nd May 2015 03:00How did the "Oxford Street of South London" come to be one of the poorest shopping streets in modern London? Read more: The Secret History Of Our Streets  Deptford High StreetSecret History Of... Deptford High St
Available until Friday, 22nd May 2015 03:00Thinking Allowed: The Ethnography Award 'Shortlist' 2015
Available until Friday, 15th April 2016 10:30A History of Ideas  Descartes Cogito Ergo Sum
Available until Thursday, 14th April 2016 08:30 
Get Started menu item
Welcome to OpenLearn
Get the most out of OpenLearn  Sign in
OpenLearn is the home of free learning from the UK's largest university, The Open University.
We have over 600 free courses available for you to study right now. More about this siteInspire me
OU on the BBC: Andrew Marr's History of the World  Survival
Andrew Marr sets off on an epic journey through 70,000 years of human history, starting with our... Read more: OU on the BBC: Andrew Marr's History of the World  SurvivalTake the photographic memory test
Can you capture scenes just by looking at them? Find out with our photographic memory test. Launch now: Take the photographic memory testFree courses
Eating to win: Activity, diet and weight control
What should we eat before and after exercise? This free course, Eating to win: activity, diet and... Try: Eating to win: Activity, diet and weight control nowSucceed with maths – Part 1
[BETA] If you feel that maths is a mystery that you want to unravel then this short 8week course... Try: Succeed with maths – Part 1 nowBecome an OU student
Learning to learn
Try our free course and get the most out of all our free courses Try: Learning to learn now
 You are here:
 Home
 Science, Maths & Technology
 Mathematics and Statistics
 Vectors and conics
 4.6 Hyperbola (e > 1)
Vectors and conics
Attempts to answer problems in areas as diverse as science, technology and economics...
Attempts to answer problems in areas as diverse as science, technology and economics involve solving simultaneous linear equations. In this unit 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.
By the end of this unit you should be able to:
 Section 1
 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 oneone correspondence between the set of points in threedimensional space and the set of ordered triples of real numbers;
 recognise the equation of a plane in three dimensions.
 Section 2
 explain what are meant by a vector, a scalar multiple of a vector, and the sum and difference of two vectors;
 represent vectors in and in terms of their components, and use components in vector arithmetic;
 use the Section Formula for the position vector of a point dividing a line segment in a given ratio;
 determine the equation of a line in or in terms of vectors.
 Section 3
 explain what is meant by the dot product of two vectors;
 use the dot product to find the angle between two vectors and the projection of one vector onto another;
 determine the equation of a plane in , given a point in the plane and the direction of a normal to the plane.
 Section 4
 explain the term conic section;
 determine the equation of a circle, given its centre and radius, and the centre and radius of a circle, given its equation;
 explain the focus–directrix definitions of the nondegenerate conics.
 Duration 20 hours
 Updated Wednesday 18th May 2011
 Intermediate level
 Posted under Mathematics and Statistics
Contents
 Introduction
 Learning outcomes
 1 Coordinate geometry: points, planes and lines
 1.1 Points, lines and distances in twodimensional Euclidean space
 1.2 Lines
 1.3 Parallel and perpendicular lines
 1.4 Intersection of two lines
 1.5 Distance between two points in the plane
 1.6 Points, planes, lines and distances in threedimensional Euclidean space
 1.7 Planes in threedimensional Euclidean space
 1.8 Intersection of two planes
 1.9 Distance between points in threedimensional Euclidean space
 1.10 Further exercises
 2 Vectors
 3 Dot product
 4 Conics
 Acknowledgements
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
the focus F lies on the xaxis, and has coordinates (ae, 0), where a > 0;
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 FP^{2} = e^{2} × PM^{2}; that is,
Multiplying out the brackets, we obtain
x^{2} − 2aex + a_{2}e_{2} + y_{2} = e_{2}x_{2} − 2aex + a_{2},
which simplifies to the equation
x_{2}(e_{2} − 1) − y_{2} = a_{2}(e_{2} − 1),
that is,
Substituting b for , so that b^{2} = a^{2}(e^{2} − 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 xaxis 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, sec^{2}t = 1 + tan^{2}t.
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).
Exercise 57
Let P be a point , t ∈ , on the hyperbola with equation x^{2} − 2y^{2} = 1.
(a) Determine the foci F and F′ of the hyperbola.
(b) Determine the gradients of FP and F′P, when these lines are not parallel to the yaxis.
(c) Find the point P on the hyperbola, in the first quadrant, for which FP is perpendicular to F′P.
Answer
(a) This hyperbola is of the form with a = 1 and , so . If e denotes the eccentricity of the hyperbola, so that b^{2} = a^{2}(e^{2} − 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 yaxis.
Similarly, the gradient of F′P is

where we know that , since F′P is not parallel to the yaxis.
(c) When FP is perpendicular to F′P, we have

We may rewrite this in the form

so 2sec^{2}t − 3 + tan^{2}t = 0.
Since sec^{2}t = 1 + tan^{2}t, it follows that 3 tan^{2}t = 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 .
About this site
OpenLearn: free learning from The Open University