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- 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 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.
- 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 focusâ€“directrix definitions of the non-degenerate 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 two-dimensional 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 three-dimensional Euclidean space
- 1.7 Planes in three-dimensional Euclidean space
- 1.8 Intersection of two planes
- 1.9 Distance between points in three-dimensional 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*x*-axis, 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} − 2*aex* + *a*_{2}*e*_{2} + *y*_{2} = *e*_{2}*x*_{2} − 2*aex* + *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 *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, 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* = *a*cosh*t*, y = *b*sinh*t* (*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} − 2*y*^{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*y*-axis.**(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 haveit 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*iswhere we know that , since

*FP*is not parallel to the*y*-axis.Similarly, the gradient of

*F*′*P*iswhere we know that , since

*F*′*P*is not parallel to the*y*-axis.**(c)**When*FP*is perpendicular to*F*′*P*, we haveWe 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 .