maths in focus - margaret grove - ch10

TERMINOLOGY

10 Locus and the Parabola

Axis: A line around which a curve is refl ected e.g. the axis of symmetry of a parabola

Chord: An interval joining any two points on a curve. In this chapter, any two points on a parabola

Circle: The locus of a point moving so that it is equidistant from a fi xed point on a plane surface

Directrix: A fi xed line from which all points equidistant from this line and a fi xed point called the focus form a parabola

Focal chord: A chord that passes through the focus

Focal length: The distance between the focus and the vertex of a parabola or the shortest distance between the vertex and the directrix

Focus: A fi xed point from which all points equidistant from this point and the directrix form a parabola

Latus rectum: A focal chord that is perpendicular to the axis of the parabola

Locus: The path traced out by a point that moves according to a particular pattern or rule. Locus can be described algebraically or geometrically

Parabola: The locus of a point moving so that it is equidistant from a fi xed point called the focus and a fi xed line called the directrix

Tangent: A straight line that touches a curve at a single point only

Vertex: The turning point (maximum or minimum point) of a parabola. It is the point where the parabola meets the axis of symmetry

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485Chapter 10 Locus and the Parabola

INTRODUCTION

THIS CHAPTER EXPANDS THE work on functions that you have already learned. It shows a method of fi nding the equation of a locus. In particular, you will study the circle and the parabola , defi ned as a locus.

DID YOU KNOW?

Locus problems have been studied since very early times. Apollonius of Perga (262–190 BC), a contemporary (and rival) of Archimedes , studied the locus of various fi gures. In his Plane Loci , he described the locus points whose ratio from two fi xed points is constant. This locus is called the ‘Circle of Apollonius’.

Apollonius also used the equation y lx2= for the parabola.

René Descartes (1596–1650) was another mathematician who tried to solve locus problems. His study of these led him to develop analytical (coordinate) geometry.

Locus

A relation can be described in two different ways. It can be a set of points that obey certain conditions, or a single point that moves along a path according to certain conditions.

A locus is the term used to describe the path of a single moving point that obeys certain conditions.

Circle

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486 Maths In Focus Mathematics Preliminary Course

EXAMPLES

Describe the locus of the following. 1. A pencil on the end of compasses.

Solution

The path of the pencil is a circle with centre at the point of the compasses.

2. A person going up an escalator (standing still on one step).

Solution

The body travels along a straight line parallel to the escalator.

3. A doorknob on a closing door.

What would the locus be if the person walks up the escalator?

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Solution

If the door could swing right around it would follow a circle. So a door closing swings through an arc of a circle.

4. A point on the number line that is 3 units from 0.

Solution

The locus is .3!

5. A point in the number plane that moves so that it is always 3 units from the y -axis.

Solution

The locus is 2 vertical lines with equations .x 3!=

Class Discussion

Describe the path of a person abseiling down a cliff.

1. a racing car driving around a track

2. a person climbing a ladder

3. a child on a swing

4. a ball’s fl ight when thrown

5. a person driving up to the 5th fl oor of a car park

10.1 Exercises

Describe the locus of the following:

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6. a point that moves along the number line such that it is always less than 2 units from 0

7. a point on the number plane that moves so that it is always 2 units from the origin

8. a point that moves so that it is always 1 unit from the x -axis

9. a point that moves so that it is always 5 units from the y -axis

10. a point that moves so that it is always 2 units above the x -axis

11. a point that moves so that it is always 1 unit from the origin

12. a point that moves so that it is always 4 units from the point ,1 2-^ h

13. a point that is always 5 units below the x -axis

14. a point that is always 3 units away from the point (1, 1)

15. a point that is always 7 units to the left of the y -axis

16. a point that is always 3 units to the right of the y -axis

17. a point that is always 8 units from the x -axis

18. a point that is always 4 units from the y -axis

19. a point that is always 6 units from the point ( 2- , 4)

20. a point that is always 1 unit from the point ( 4- , 5).

A locus describes a single point ,P x y^ h that moves along a certain path. The equation of a locus can often be found by using ,P x y^ h together with the information given about the locus.

EXAMPLES

1. Find the equation of the locus of a point ,P x y^ h that moves so that it is always 3 units from the origin.

Solution

You may recognise this locus as a circle, centre ,0 0^ h radius 3 units. Its equation is given by 9.x y2 2+ =

Alternatively, use the distance formula.

d x x y y

d x x y yor2 1

22 1

2

22 1

22 1

2

= - + -

= - + -

_ __ _

i ii i

You studied this formula in Chapter 7. It is easier to use d

2

than d to fi nd the equation of the locus.

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Let ,P x y^ h be a point of the locus.

0

PO

PO

x y

x y

3

9

0 9

9

We want

i.e. 2

2 2

=

=

- + - =

+ =

22^ ^h h

2. Find the equation of the locus of point ,x yP ^ h that moves so that distance PA to distance PB is in the ratio 2:1 where , , .A B3 1 2 2and= - = -^ ^h h

Solution


: :

[ 3 ] 2 [ 2 ]3 [ 2 ]

4 4 4 4

PA PB

PBPA

PA PB

PA PB

PB

x y x yx y x y

x x y y x x y x

x x y y

x x y y

x x y y

2 1

12

2

2

4

1 41 4 2

6 9 2 1 4

4 16 16 4 16 16

0 3 22 3 18 22

3 22 3 18 22 0

i.e.

i.e.

or

2

2

2 2 2

2

2 2 2 2

2 2

2 2

2 2

`

=

=

=

=

=

- - + - = - + - -

+ + - = - + +

+ + + - + = - + + + +

= - + + + +

= - + + +

- + + + =

2

2

2 2 2

]

^ ^ ^ ^^ ^ ^ ^

^

g

h h h hh h h h

h

$ .

3. Find the equation of the locus of a point ,x yP ^ h that moves so that the line PA is perpendicular to line PB , where ,A 1 2= ^ h and , .B 3 1= - -^ h

Place P anywhere on the number plane.

This is the equation of a circle.

Use the distance formula as in

Example 1.

CONTINUED

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Solution

Let ,x yP ^ h be a point of the locus. For perpendicular lines, m m 11 2 = -

:

:

m x xy y

PA mx

y

PB m

x

yx

y1

2

3

1

3

1

Using

1

2

2 1

2 1=

-

-

=-

-

=-

-

=+

+

-

-

]]gg

For PA perpendicular to PB

x

y

x

y

x x

y y

y y x x

x x

x x y y

1

2

3

11

2 3

21

2 2 3

2 3

2 5 0i.e.

2

2

2 2

2

2 2

#-

-

+

+= -

+ -

- -= -

- - = - + -

= - - +

+ + - - =

^ h

4. Find the equation of the locus of point ,P x y^ h that is equidistant from fi xed point ,A 1 2-^ h and fi xed line with equation 5.y =

Solution

Let ,P x y^ h be a point of the locus. B has coordinates , .x 5^ h

[ ]

PA PB

PA PB

x y x x y

x y y

x x y y y y

x x y

1 2 5

1 2 5

2 1 4 4 10 25

2 14 20 0

We want

i.e. 2 2

2

2 2 2

2

=

=

- + - - = - + -

- + + = -

- + + + + = - +

- + - =

2 2 2

2 2 2

^ ^ ^ ^^ ^ ^h h h hh h h

These results come from Chapter 7.

The locus is a circle with diameter AB .

This is the equation of a parabola. Can you see where the parabola lies?

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1. Find the equation of the locus of point ,P x y^ h that moves so that it is always 1 unit from the origin.

2. Find the equation of the locus of point ,P x y^ h that moves so that it is always 9 units from the point , .1 1- -^ h

3. Find the equation of the locus of a point that moves so that it is always 2 units from the point , .5 2-^ h

4. Find the equation of the locus of point ,P x y^ h that moves so that it is equidistant from the points ,3 2^ h and , .1 5-^ h

5. Find the equation of the locus of a point that moves so that it is equidistant from the points ,4 6-^ h and , .2 7-^ h

6. Find the equation of the locus of point ,P x y^ h that moves so that it is equidistant from the x -axis and the y -axis.

7. Find the equation of the locus of a point P that moves so that PA is twice the distance of PB where ,A 0 3= ^ h and , .B 4 7= ^ h

8. Find the equation of the locus of point ,P x y^ h that moves so that the ratio of PA to PB is :3 2 where ,A 6 5= -^ h and , .B 3 1= -^ h

9. Find the equation of the locus of a point that moves so that it is equidistant from the point ,2 3-^ h and the line 7.y =

10. Find the equation of the locus of a point that moves so that it is equidistant from the point ,0 5^ h and the line 5.y = -

11. Find the equation of the locus of a point that moves so that it is equidistant from the point ,2 0^ h and the line 6.x =



14. Find the equation of the locus of a point ,P x y^ h that moves so that the line PA is perpendicular to line PB where ,A 1 3= -^ h and , .B 4 5= ^ h

15. Find the equation of the locus of a point ,P x y^ h that moves so that the line PA is perpendicular to line PB , where ,A 4 0= -^ h and , .B 1 1= ^ h

16. Find the equation of the locus of a point ,P x y^ h that moves so that the line PA is perpendicular to line PB where ,A 1 5= ^ h and , .B 2 3= - -^ h

17. Point P moves so that 4PA PB2 2+ = where ,A 3 1= -^ h and , .B 5 4= -^ h Find the equation of the locus of P .

18. Point P moves so that 12PA PB2 2+ = where ,A 2 5= - -^ h and , .B 1 3= ^ h Find the equation of the locus of P .

19. Find the equation of the locus of a point that moves so that its distance from the line 3 4 5 0x y+ + = is always 4 units.

10.2 Exercises

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20. Find the equation of the locus of a point that moves so that its distance from the line 12 5 1 0x y- - = is always 1 unit.

21. Find the equation, in exact form, of the locus of a point that moves so that its distance from the line 2 3 0x y- - = is always 5 units.

22. Find the equation of the locus of a point that moves so that it is equidistant from the line 4 3 2 0x y- + = and the line 3 4 7 0.x y+ - =

23. Find the equation of the locus of a point that moves so that it is equidistant from the line 3 4 5 0x y+ - = and the line 5 12 1 0.x y+ - =

24. Given two points ,A 3 2-^ h and , ,B 1 7-^ h fi nd the equation of the locus of ,P x y^ h if the gradient of PA is twice the gradient of PB .

25. If R is the fi xed point ,3 2^ h and P is a movable point , ,x y^ h fi nd the equation of the locus of P if the distance PR is twice the distance from P to the line .y 1= -

PROBLEM

Can you see 2 mistakes in the solution to this question ? Find the locus of point ,P x y^ h that moves so that its perpendicular distance from the line 12 5 1 0x y+ - = is always 3 units.

Solution


| |

| |

| |

| |

| |

da b

ax by c

x y

x y

x y

x y

x y

x y

35 12

5 12 1

25 144

5 12 1

169

5 12 1

13

5 12 1

39 5 12 1

0 5 12 40

2 2

1 1

2 2

`

=+

+ +

=+

+ -

=+

+ -

=+ -

=+ -

= + -

= + -

Can you fi nd the correct locus?

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Circle as a Locus

The locus of point P (x, y) that is always a constant distance from a fi xed point is a circle.

The circle, centre ,0 0^ h and radius r , has the equation x y r2 2 2+ =

Proof

Find the equation of the locus of point ,P x y^ h that is always r units from the origin.

Let ,x yP ^ h be a point of the locus.

0

OP r

OP r

x y r

x y r

0

i.e. 2 2

2

2 2 2

=

=

- + - =

+ =

22^ ^h h

So x y r2 2 2+ = is the equation of the locus. It describes a circle with radius r and centre , .0 0^ h

The circle, centre ,a b^ h and radius r , has the equation

x a y b r2- + - =22^ ^h h

Proof

Find the equation of the locus of point ,P x y^ h that is always r units from point , .A a b^ h

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AP r

AP r

x a y b r

i.e. 2 2

2

=

=

- + - =22^ ^h h

So x a y b r2 2 2- + - =] ^g h is the equation of the locus. It describes a circle with radius r and centre , .a b^ h

EXAMPLES

1. Find the equation of the locus of a point that is always 2 units from the point , .1 0-^ h

Solution

This is a circle with radius 2 and centre , .1 0-^ h Its equation is in the form

[ 1 ]

1

x a y b r

x y

x y

x x y

x x y

0 2

4

2 1 4

2 3 0

i.e.

2

2 2 2

2

2 2

2 2

- + - =

- - + - =

+ + =

+ + + =

+ + - =

2 2

2

^ ^^ ^^

h hh h

h

2. Find the radius and the coordinates of the centre of the circle .x x y y2 6 15 02 2+ + - - =

Solution

We put the equation into the form .x a y b r2- + - =2 2^ ^h h To do this we complete the square. In general, to complete the square on ,x bx2 + add

2b 2c m to give:

2 2

x bx b x b22 2

+ + = +c cm m

First we move any constants to the other side of the equation, then complete the square. To complete the square on 2 ,x x2 + we add .

22 1

2

=c m

You could fi nd this equation by using (x, y)P and treating the question as a locus problem.

You learned how to complete the square in Chapter 3.

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To complete the square on ,y y62 - we add .26 9

2

=c m x x y y2 6 15 02 2+ + =- -

2x x2 + y y62+ - 15= 2 6 151 9 1 9x x y y2 2+ + - =+ + + +

1 25x y 3+ + - =22^ ^h h x y1 3 52- - + - =22]^ ^gh h

The equation is in the form .x a y b r2- + - =2 2^ ^h h This is a circle, centre ,1 3-^ h and radius 5.

1. Find the length of the radius and the coordinates of the centre of each circle.

(a) 100x y2 2+ = (b) 5x y2 2+ = (c) x y4 5 16- + - =22^ ^h h (d) x y5 6 49- + + =22^ ^h h (e) x y 3 812 2+ - =^ h

2. Find the equation of each circle in expanded form (without grouping symbols) .

Centre (0, 0) and radius 4 (a) Centre (3, 2) and radius 5 (b) Centre (c) ,1 5-^ h and radius 3 Centre (2, 3) and radius 6 (d) Centre (e) ,4 2-^ h and radius 5 Centre (f) ,0 2-^ h and radius 1 Centre (4, 2) and radius 7 (g) Centre (h) ,3 4- -^ h and radius 9 Centre (i) ,2 0-^ h and radius 5 Centre (j) ,4 7- -^ h and

radius .3

3. Find the equation of the locus of a point moving so that it is 1 unit from the point , .9 4-^ h

4. Find the equation of the locus of a point moving so that it is 4 units from the point , .2 2- -^ h

5. Find the equation of the locus of a point moving so that it is 7 units from the point , .1 0^ h

6. Find the equation of the locus of a point moving so that it is 2 units from the point , .3 8-^ h

7. Find the equation of the locus of a point moving so that it is 2 units from the point , .5 2-^ h

8. Find the equation of a circle with centre ,0 0^ h and radius 3 units.

9. Find the equation of a circle with centre ,1 5^ h and radius 1 unit.

10. Find the equation of a circle with centre ,6 1-^ h and radius 6 units.

11. Find the equation of a circle with centre ,4 3^ h and radius 3 units.

12. Find the equation of a circle with centre ,0 3-^ h and radius 2 2 units.

13. Find the coordinates of the centre and the length of the radius of each circle.

(a) x x y y4 2 4 02 2+ =- - - (b) x x y y8 4 5 02 2+ + =- - (c) x y y2 02 2+ =-

10.3 Exercises

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Concentric circles have the same centre.

(d) x x y y10 6 2 02 2+ + =- - (e) x x y y2 2 1 02 2+ + + =- (f) x x y12 02 2+ =- (g) x x y y6 8 02 2+ + =- (h) x x y y20 4 40 02 2+ + + =- (i) x x y y14 2 25 02 2+ + + =- (j) x x y y2 4 5 02 2+ + + - =

14. Find the centre and radius of the circle with equation given by 6 2 6 0.x x y y2 2- + + - =

15. Find the centre and radius of the circle with equation given by 4 10 4 0.x x y y2 2- + - + =

16. Find the centre and radius of the circle with equation given by .x x y y2 12 12 02 2+ + + - =

17. Find the centre and radius of the circle with equation given by 8 14 1 0.x x y y2 2- + - + =

18. Find the centre and radius of the circle with equation given by 3 2 3 0.x x y y2 2+ + - - =

19. Sketch the circle whose equation is given by 4 2 1 0.x x y y2 2+ + - + =

20. Prove that the line 3 4 21 0x y+ + = is a tangent to the circle 8 4 5 0.x x y y2 2- + + - =

21. (a) Show that 2 4 1 0x x y y2 2- + + + = and 2 4 4 0x x y y2 2- + + - = are concentric .

Find the difference between (b) their radii.

22. Given two points ,A 2 5-^ h and , ,B 4 3-^ h fi nd the equation of the circle with diameter AB .

23. Find the exact length of AB where A and B are the centres of the circles 6 0x x y2 2- + = and 4 6 3 0x x y y2 2+ + + - = respectively.

24. (a) Find the length of XY where X and Y are the centres of the circles 6 2 1 0x x y y2 2+ + - + = and 4 2 1 0x x y y2 2- + - + = respectively.

Find the radius of each circle. (b) What conclusion can you draw (c)

from the results for (a) and (b)?

25. Show that the circles 4x y2 2+ = and 2 4 4 0x x y y2 2+ + - - = both have 3 4 10 0x y+ + = as a tangent.

26. A circle has centre ,C 1 3-^ h and radius 5 units.

Find the equation of the (a) circle.

The line (b) 3 1 0x y- + = meets the circle at two points. Find their coordinates.

Let the coordinates be (c) X and Y , where Y is the coordinate directly below the centre C . Find the coordinates of point Z , where YZ is a diameter of the circle.

Hence show (d) .ZXY 90c+ =

27. (a) Find the perpendicular distance from ,P 2 5-^ h to the line 5 12 2 0.x y+ - =

Hence fi nd the equation (b) of the circle with centre P and tangent .x y5 12 2 0+ - =

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Parabola as a Locus

The locus of a point that is equidistant from a fi xed point and a fi xed line is always a parabola. The fi xed point is called the focus and the fi xed line is called the directrix.

Work on the parabola as a locus is very important, as the properties of the parabola are useful to us. The parabola is used in lenses of glasses and cameras, in car headlights, and for bridges and radio telescope dishes.

DID YOU KNOW?

Any rope or chain supporting a load (e.g. a suspension bridge) is in the shape of a parabola.

Find some examples of suspension bridges that have a parabola shaped chain.

Other bridges have ropes or chains hanging freely. These are not in the shape of a parabola, but are in a shape called a catenary. Can you fi nd some bridges with this shape?

More recent bridges are cable-stayed, where ropes or chains are attached to towers, or pylons, and fan out along the sides of the bridge. An example is the Anzac Bridge in Sydney.

There are many different bridge designs. One famous bridge in Australia is the Sydney Harbour Bridge.

Research different bridge designs and see if you can fi nd some with parabolic shapes.

Parabola with vertex at the origin

Just as the circle has a special equation when its centre is at the origin, the parabola has a special equation when its vertex is at the origin. Both also have a more general formula.

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The locus of a point that is equidistant from a fi xed point and a fi xed line is always in the shape of a parabola.

If the fi xed point is (0, a ) and the fi xed line is y a= - (where a 02 ), then one of the equidistant points is the origin (0, 0). The distance between the points (0, 0) and (0, a ) is a units.

The point on y a= - directly below the origin is , a0 -^ h and the distance from (0, 0) to , a0 -^ h is also a units.

(0, a)

(0,-a) y=-a

a

a

y

x

To fi nd the equation of the parabola, we use the general process to fi nd the equation of any locus. The features of the parabola have special names.

A parabola is equidistant from a fi xed point and a fi xed line. The fi xed point is called the • focus. The fi xed line is called the • directrix. The turning point of the parabola is called the • vertex. The axis of symmetry of the parabola is called its • axis. The distance between the vertex and the focus is called the • focal length. An interval joining any two points on the parabola is called a • chord. A chord that passes through the focus is called a • focal chord. The focal chord that is perpendicular to the axis is called the • latus rectum. A • tangent is a straight line that touches the parabola at a single point.

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The locus of point ,P x y^ h moving so that it is equidistant from the point , a0^ h and the line y a= - is a parabola with equation

4x ay2 =

PARABOLA x2 4= ay

The parabola 4x ay2 = has • focus at , a0^ h • directrix with equation y a= - • vertex at ,0 0^ h • axis with equation 0x = focal length• the distance from the vertex to the focus with length a latus rectum• that is a horizontal focal chord with length 4 a

Since the focal length is a , a is always a positive number.

Proof

Let ,P x y^ h be a point of the locus. Taking the perpendicular distance from P to the line ,y a= - point , .B x a= -^ h

0 [ ]

PA PB

PA PB

x y a x x y a

x y a y a

x y ay a y ay a

x ay

2 2

4

2 2

2

2

2 2 2 2 2

2

`

=

=

- + - = - + - -

+ - = +

+ - + = + +

=

2 2 2

2 2

^ ^ ^ ^^ ^

h h h hh h

Class Investigation

Find the equation of the locus if point ,P x y^ h is equidistant from , a0 -^ h and .y a=

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EXAMPLES

1. Find the equation of the parabola whose focus has coordinates ,0 2^ h and whose directrix has equation 2.y = -

Solution

The focus has coordinates in the form , a0^ h and the directrix has equation in the form ,y a= - where 2.a = ` the parabola is in the form 4x ay2 = where 2a =

( )x y

x y

4 2

8

i.e. 2

2

=

=

2. Find the coordinates of the focus and the equation of the directrix of (a)

the parabola .x y202 = Find the points on the parabola at the endpoints of the latus rectum (b)

and fi nd its length.

Solution

The parabola (a) 20x y2 = is in the form 4x ay2 =

4 20

5

a

a`

=

=

The focal length is 5 units. We can fi nd the coordinates of the focus and the equation of the directrix in two ways.

Method 1: Draw the graph 20x y2 = and count 5 units up and down from the origin as shown.

y

x

(0, 5)

(0, -5) y=-5

x2=20y

5

5

The focus is (0, 5) and the directrix has equation y = -5.

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Method 2: The focus is in the form (0, a ) where .a 5= So the focus is (0, 5).

The directrix is in the form y a= - where .a 5= So the directrix is .y 5= -

The latus rectum is a focal chord that is perpendicular to the axis of (b) the parabola as shown

(0, 5)

x2= 20 y

y

x

The endpoints of the latus rectum will be where the line 5y = and the parabola intersect. Substitute 5y = into the parabola.

x y

x

2020 5100

10010

2

!

!

=

=

=

=

=

] g

So the endpoints are ( 10- , 5) and (10, 5).

(0, 5) (10, 5)(-10, 5)

y

x

x2=20 y

From the graph, the length of the latus rectum is 20 units.

CONTINUED

The latus rectum is 4 a units long which gives

20 units.

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3. Find the equation of the focal chord to the parabola x y42 = that passes through ( 4- , 4).

Solution

The parabola 4x y2 = is in the form .x ay42 =

4 4

1

a

a`

=

=

The focal length is 1 unit. The focus is 1 unit up from the origin at (0, 1) and the focal chord also passes through ( 4- , 4).

(0, 1)

(-4, 4)

y

x

x2=4y

We can fi nd the equation of the line between (0, 1) and ( 4- , 4) by using either formula

y y m x x x xy y

x xy y

or1 11

1

2 1

2 1- = -

-

-=

-

-_ i

x xy y

x xy y

x

y

xy

y xy x

x y

0

1

4 04 1

14

3

4 1 34 4 3

0 3 4 4

1

1

2 1

2 1

-

-=

-

-

-

-=

- -

-

-=

-

- - =

- + =

= + -

^ h

You used these formulae in Chapter 7.

As you saw in the previous chapter, a parabola can be concave downwards. Can you guess what the equation of this parabola might be?

PARABOLA x2 4= - ay

The locus of a point P ( x , y ) moving so that it is equidistant from the point , a0 -^ h and the line y a= is a parabola with equation 4x ay2 = -

ch10.indd 502 7/18/09 3:50:07 PM


Proof

A(0, -a)

P(x, y)

B(x, a) y=a

y

x

Let P ( x , y ) be a point of the locus. Taking the perpendicular distance from P to the line ,y a= point , .B x a= ^ h

PA PB

PA PB

x y a x x y ax y a y a

x y ay a y ay a

x ay

0

2 2

4

2 2

2 2

2 2 2

2 2 2 2 2

2

`

=

=

- + - - = - + -

+ + = -

+ + + = - +

= -

22^ ^ ^ ^^ ^

h h h hh h

7 A

The parabola 4x ay2 = - has • focus at , a0 -^ h • directrix with equation y a= vertex • at (0, 0) axis• with equation 0x = focal length• a latus rectum• a horizontal focal chord with length 4 a

ch10.indd 503 7/18/09 3:51:28 PM


EXAMPLES

1. Find the equation of the parabola with focus ,0 4-^ h and directrix .y 4=

Solution

If we draw this information, the focus is below the directrix as shown. So the parabola will be concave downwards (the parabola always turns away from the directrix).

y

x

(0, -4)

y=4

4

4

The focal length is 4 so .a 4= The parabola is in the form 4x ay2 = - where .a 4=

4416

x ayy

y4

2 = -

= -

= -

] g

2. Find the coordinates of the vertex, the coordinates of the focus and the equation of the directrix of the parabola .x y122 = -

Solution

The parabola x y122 = - is in the form .x ay42 = -

a 3` =

a4 12=

The focal length is 3 units. The vertex is (0, 0). We can fi nd the coordinates of the focus and the equation of the directrix in two ways. Method 1: Draw the graph x y122 = - and count 3 units up and down from the origin as shown. (The parabola is concave downward.)

ch10.indd 504 7/18/09 3:51:30 PM


y

x

(0, -3)

y=3

3

3

x2=-12y

Counting down 3 units, the focus is , .0 3-^ h Counting up 3 units, the directrix has equation .y 3= Method 2: The focus is in the form , a0 -^ h where .a 3= So the focus is , .0 3-^ h The directrix is in the form y a= where .a 3= So the directrix is .y 3=

3. Find the equation of the parabola with focal length 5 and whose vertex is ,0 0^ h and equation of the axis is 0.x =

Solution

Vertex ,0 0^ h and axis given by 0x = give a parabola in the form ,x ay42

!= since there is not enough information to tell whether it is concave upwards or downwards. This gives two possible parabolas.

CONTINUED

ch10.indd 505 7/18/09 3:51:31 PM


( )

a

x y

x y

5

4 5

20

Focal length of 5 means

The equation is

i.e.

2

2

!

!

=

=

=

1. Find the equation of each parabola.

focus (0, 5), directrix (a) 5y = - focus (0, 9), directrix (b) 9y = - focus (0, 1), directrix (c) y 1= - focus (0, 4), directrix (d) 4y = - focus (0, 10), directrix (e)

y 10= - focus (0, 3), directrix (f) 3y = - focus (0, 6), directrix (g) 6y = - focus (0, 11), directrix (h)

y 11= - focus (0, 2), directrix (i) 2y = - focus (0, 12), directrix (j)

y 12= -


focus (0, (a) 1- ), directrix 1y = focus (0, (b) 3- ), directrix 3y = focus (0, (c) 4- ), directrix 4y = focus (0, (d) 7- ), directrix 7y = focus (0, (e) 6- ), directrix 6y = focus (0, (f) 9- ), directrix 9y = focus (0, (g) 8- ), directrix 8y = focus (0, (h) 2- ), directrix 2y = focus (0, (i) 15- ), directrix

15y = focus (0, (j) 13- ), directrix

y 13=

3. Find(i) the coordinates of the focus and(ii) the equation of the directrix of

(a) x y42 = (b) 28x y2 = (c) x y162 =

(d) 36x y2 = (e) 40x y2 = (f) 44x y2 = (g) x y122 = (h) 6x y2 = (i) x y102 = (j) x y152 =

4. Find (i) the coordinates of the focus and (ii) the equation of the directrix of

(a) 4x y2 = - (b) 24x y2 = - (c) 8x y2 = - (d) 48x y2 = - (e) 20x y2 = - (f) 16x y2 = - (g) 32x y2 = - (h) 40x y2 = - (i) 2x y2 = - (j) x y222 = -

5. Find the equation of the parabola with

coordinates of the focus (a) ,0 7^ h and equation of the directrix y 7= -

coordinates of the focus (b) ,0 11^ h and equation of the directrix y 11= -

coordinates of the focus (c) ,0 6-^ h and equation of the directrix y 6=

coordinates of the focus (d) ,0 2^ h and coordinates of the vertex , .0 0^ h

10.4 Exercises

ch10.indd 506 8/4/09 11:42:11 AM


coordinates of the vertex (e) , ,0 0^ h equation of the axis 0x = and focal length 3

coordinates of the vertex (f) , ,0 0^ h equation of the axis 0x = and focal length 8

coordinates of the vertex (g) ,0 0^ h and equation of the axis 0,x = and passing through the point ,8 2-^ h

coordinates of the vertex (h) ,0 0^ h and equation of the axis 0,x = and passing through the point , .1 7-^ h

6. Find the coordinates of the focus, the equation of the directrix and the focal length of the parabola

(a) 8x y2 = (b) 24x y2 = (c) x y122 = - (d) 2x y2 = (e) 7x y2 = - (f) x y2 2 =

7. Find the equation of the focal chord that cuts the curve 8x y2 = at , .4 2-^ h

8. The tangent with equation 2 4 0x y- - = touches the parabola 4x y2 = at A . Find the coordinates of A .

9. The focal chord that cuts the parabola x y62 = - at ,6 6-^ h cuts the parabola again at X . Find the coordinates of X .

10. Find the coordinates of the endpoints of the latus rectum of the parabola 8 .x y2 = - What is the length of the latus rectum?

11. The equation of the latus rectum of a parabola is given by 3.y = - The axis of the parabola is 0,x = and its vertex is , .0 0^ h

Find the equation of the (a) parabola.

Find the equation of the (b) directrix.

Find the length of the focal (c) chord that meets the parabola at

, .231

-c m

12. (a) Show that the point ,3 3-^ h lies on the parabola with equation 3 .x y2 =

Find the equation of the line (b) passing through P and the focus F of the parabola.

Find the coordinates of the (c) point R where the line PF meets the directrix.

13. (a) Find the equation of chord

PQ where ,P 141

-c m and ,Q 2 1^ h lie on the parabola 4 .x y2 =

Show that (b) PQ is not a focal chord.

Find the equation of the circle (c) with centre Q and radius 2 units.

Show that this circle passes (d) through the focus of the parabola.

14. (a) Show that ,Q aq aq2 2_ i lies on the parabola 4 .x ay2 =

Find the equation of the focal (b) chord through Q .

Prove that the length of the (c) latus rectum is 4 a .

ch10.indd 507 7/18/09 3:51:32 PM


Investigation

Sketch the parabola .x y2= You may like to complete the table below to help you with its sketch.

x

y 3- 2- 1- 0 1 2 3

Is this parabola a function? What is its axis of symmetry?

The parabola that has y 2 rather than x 2 in its equation is a sideways parabola. It still has the same properties, but generally the x and y values are swapped around.

PARABOLA y2 4= ax

The locus of point ,P x y^ h moving so that it is equidistant from the point ,a 0^ h and the line x a= - is a parabola with equation

4y ax2 =

Proof

Find the equation of the locus of point , ,P x y^ h which moves so that it is equidistant from the point ,a 0^ h and the line .x a= -

Coordinates of B are , .a y-^ h

[ ]

PA PB

PA PB

x a y x a y y

x a y x a

x ax a y x ax a

y ax

0

2 2

4

We want

i.e. 2 2

2

2

2 2 2 2 2

2

=

=

- + - = - - + -

- + =

- + + = + +

=

+

2 2 2 2

2 2

^ ^ ^ ^^ ^h h h h

h h

ch10.indd 508 7/18/09 3:51:33 PM


The parabola y ax42 = has • focus at ,a 0^ h equation of • directrix x a= - vertex• at ,0 0^ h axis• with equation 0y = focal length• the distance from the vertex to the focus with length a • latus rectum that is a vertical focal chord with length 4 a

EXAMPLES

1. Find the equation of the parabola with focus (7, 0) and directrix .x 7= -

Solution

If we draw this information, the focus is to the right of the directrix as shown (the parabola always turns away from the directrix). So the parabola turns to the right.

y

x

x=-7

77

(7, 0)

CONTINUED

ch10.indd 509 7/18/09 3:51:34 PM


The focal length is 7 so .a 7= The parabola is in the form 4y ax2 = where .a 7=

.

y axx

x

44 728

2 =

=

=

^ h

2. Find the coordinates of the focus and the equation of the directrix of the parabola .y x322 =

Solution

The parabola 32y x2 = is in the form .y ax42 =

4 32

8

a

a`

=

=

The focal length is 8 units. Method 1: Draw the graph 32y x2 = and count 8 units to the left and right from the origin as shown. (The parabola turns to the right.)

y

x

x=-8

88

(8, 0)

y2=32x22

Counting 8 units to the right, the focus is (8, 0). Counting 8 units to the left, the directrix has equation .x 8= - Method 2: The focus is in the form ( a , 0) where .a 8= So the focus is (8, 0). The directrix is in the form x a= - where .a 8= So the directrix is .x 8= -

A parabola can also turn to the left.

ch10.indd 510 7/18/09 3:51:35 PM


PARABOLA y2 4= - ax

The locus of a point P ( x , y ) moving so that it is equidistant from the point ,a 0-^ h and the line x a= is a parabola with equation 4y ax2 = -

Proof

y

x

P (x, y)B (a, y)

A (- a, 0)

x=a

Let P ( x , y ) be a point of the locus. Taking the perpendicular distance from P to the line ,x a= point , .B a y= ^ h

PA PB

PA PB

x a y x a y yx a y x a

x ax a y x ax a

y ax

0

2 2

4

2 2

2 2

2

2 2 2 2 2

2

`

=

=

- - + - = - + -

+ + = -

+ + + = - +

= -

2 2

2 2

^ ^ ^ ^^ ^h h h h

h h7 A

ch10.indd 511 7/18/09 4:13:17 PM


The parabola 4y ax2 = - has • focus at ( a- , 0) • directrix with equation x a= vertex • at (0, 0) axis• with equation 0y = focal length• a latus rectum• a vertical focal chord with length 4 a

EXAMPLES

1. Find the equation of the parabola with focus ( 4- , 0) and directrix .x 4=

Solution

Drawing this information shows that the parabola turns to the left.

y

x(-4, 0)

x=4

4 4

The focal length is 4 so .a 4= The parabola is in the form 4y ax2 = - where .a 4=

.

y axx

x

44 416

2 = -

= -

= -

^ h

2. Find the coordinates of the focus and the equation of the directrix of the parabola .y x22 = -

Solution

The parabola 2y x2 = - is in the form .y ax42 = -

4 2a

a21

`

=

=

The focal length is 21 unit.

ch10.indd 512 7/18/09 3:51:37 PM


Method 1: Draw the graph 2y x2 = - and count

21 unit to the left and right from the

origin as shown. (The parabola turns to the left.)

12

12

(-12

, 0)

12

x =

y

x

Counting 21 units to the left, the focus is , .

21 0-c m

Counting 21 units to the right, the directrix has equation .x

21

=

Method 2: The focus is in the form ( a- , 0) where .a

21

=

So the focus is , .21 0-c m

The directrix is in the form x a= where .a21

=

So the directrix is .x21

=


focus (2, 0), directrix (a) 2x = - focus (5, 0), directrix (b) 5x = - focus (14, 0), directrix (c)

x 14= - focus (9, 0), directrix (d) 9x = - focus (8, 0), directrix (e) 8x = - focus (6, 0), directrix (f) 6x = - focus (7, 0), directrix (g) 7x = - focus (3, 0), directrix (h) 3x = - focus (4, 0), directrix (i) 4x = - focus (1, 0), directrix (j) x 1= -


focus ((a) 9- , 0), directrix 9x = focus ((b) 4- , 0), directrix 4x = focus ((c) 10- , 0), directrix x 10= focus ((d) 6- , 0), directrix 6x = focus ((e) 2- , 0), directrix 2x = focus ((f) 12- , 0), directrix x 12= focus ((g) 11- , 0), directrix x 11= focus ((h) 5- , 0), directrix 5x = focus ((i) 3- , 0), directrix 3x = focus ((j) 7- , 0), directrix x 7=

10.5 Exercises

ch10.indd 513 7/18/09 3:51:38 PM



(a) 8y x2 = (b) y x122 = (c) y x162 = (d) 4y x2 = (e) 28y x2 = (f) 32y x2 = (g) 24y x2 = (h) 36y x2 = (i) y x2 = (j) y x182 =


(a) 8y x2 = - (b) y x122 = - (c) 28y x2 = - (d) 4y x2 = - (e) 24y x2 = - (f) 52y x2 = - (g) 60y x2 = - (h) 2y x2 = - (i) 26y x2 = - (j) y x52 = -

5. Find the equation of the parabola with

coordinates of the focus (a) ,5 0^ h and equation of the directrix x 5= -

coordinates of the focus (b) ,1 0^ h and equation of the directrix x 1= -

coordinates of the focus (c) ,4 0-^ h and equation of the directrix x 4=

coordinates of the focus (d) ,3 0^ h and coordinates of the vertex ,0 0^ h

coordinates of the vertex (e) ,0 0^ h equation of the axis 0y = and focal length 9

coordinates of the vertex (f) , ,0 0^ h equation of the axis 0y = and focal length 2

coordinates of the vertex (g) ,0 0^ h and equation of the axis 0y = and passing through the point ,3 6^ h

coordinates of the vertex (h) ,0 0^ h and equation of the axis 0y = and passing through the point , .2 1^ h

6. Find the coordinates of the focus, the equation of the directrix and the focal length of the parabola

(a) 8y x2 = (b) 4y x2 = (c) y x122 = - (d) 6y x2 = (e) 5y x2 = - (f) y x3 2 =

7. Find the equation of the focal chord that cuts the curve y x162 = at , .4 8^ h

8. Find the length of the latus rectum of the parabola .y x122 = What are the coordinates of its endpoints?

9. The line with equation 3 27 0x y- - = meets the parabola 4y x2 = at two points. Find their coordinates.

10. Let ,R51 2-c m be a point on the

parabola 20 .y x2 = Find the equation of the focal (a)

chord passing through R . Find the coordinates of the (b)

point Q where this chord cuts the directrix.

Find the area of (c) D OFQ where O is the origin and F is the focus.

Find the perpendicular (d) distance from the chord to the point , .P 1 7- -^ h

Hence fi nd the area of (e) D PQR .

ch10.indd 514 7/31/09 5:02:24 PM


Application

A parabolic satellite dish receives its signals through the focus. If the dish has height 12 m and a span of 20 m, fi nd where the focus should be placed, to the nearest mm.

SOLUTION

The parabola is of the form 4x ay2= and passes through (10, 12) and ( 10, 12)-

Substituting (10, 12) gives

10 4 (12)

100 48

2.083

a

a

a

2=

=

=

So the focus should be placed 2.083 m from the vertex. This is 2083 mm to

the nearest millimetre.

1. 4x ay2 =

y

x

x2=4ay

Focus (0, a)

Directrix y=-a

Here is a summary of the 4 different types of parabola with the vertex at the origin.

ch10.indd 515 7/18/09 3:51:40 PM


2. 4x ay2 = -

y

x2=-4ay

Focus (0, -a)

Directrix y=a

x

3. 4y ax2 =

y

y2=4ax

Directrixx=-a

xFocus(a, 0)

4. 4y ax2 = -

y2=-4ax

Directrixx=a

y

xFocus(-a, 0)

General Parabola

When the parabola does not have its vertex at the origin, there is a more general formula.

Since we use a to mean the focal length, we cannot use ( a , b ) as the vertex. We use ( h , k ) instead.

ch10.indd 516 7/18/09 3:51:41 PM


PARABOLA (x - h)2 = 4a(y - k)

The concave upwards parabola with vertex ( h , k ) and focal length a has equation x h a y k4- = -2^ ^h h

Proof

Find the equation of the parabola with vertex ,h k^ h and focal length a .

Counting up a units from vertex V gives the focus , .F h k a= +^ h Counting down a units from V gives the point on the directrix , .D h k a= -^ h So the equation of the directrix is given by .y k a= - We fi nd the equation of the locus of ,x yP ^ h that is equidistant from point ,h k aF +^ h and line .y k a= -

B has coordinates , .x k a-^ h

i.e.

We want

[ ] [ ]

[ ] [ ]

PF PB

PF PB

x h y k a x x y k a

x h y k a y k a

x h y k a y k ay k a y k a y k a y k a

y k aay ak

a y k

2 2 24 4

4

difference of two squares

2 2

2 2

#

=

=

- + - + = - + - +

- + - - = - +

- = - + - - -

= - + + - - - + - - -

= -

= -

= -

2 2

2 2 2

2 2 2

^ ^ ^ ^^ ^ ^

^ ^ ^^ ^ ^ ^

^^ ^

^

h h h hh h h

h h hh h h h

hh h

h

ch10.indd 517 7/18/09 4:03:50 PM


The parabola x h y ka42- -=^ ^h h has • axis parallel to the y -axis • vertex at ,h k^ h focus• at ,h k a+^ h • directrix with equation y k a= -

EXAMPLES

1. Find the equation of the parabola with focus ,2 3^ h and directrix with equation 7.y = -

Solution

Coordinates of B are , .2 7-^ h The vertex is the midpoint of ,2 3^ h and , .2 7-^ h ,2 2vertex` = -^ h Focal length is the distance from the focus to the vertex. a 5` = From the diagram the parabola is concave upwards. The equation is in the form

[ ]

x h a y k

x yy

x x y

x x y

4

2 4 5 220 2

4 4 20 40

4 20 36 0

i.e.

2

2

- = -

- = - -

= +

- + = +

- - - =

2

2

^ ^^ ^ ^

^

h hh h h

h

2. Find the coordinates of the vertex and the focus, and the equation of the directrix, of the parabola with equation .x x y6 12 3 02 + - - =

Draw a diagram to fi nd the vertex and to fi nd a .

ch10.indd 518 7/18/09 4:03:51 PM


PARABOLA (x - h)2 = - 4a(y - k)

Solution

Complete the square on x .

3( )

x x y

x x y

x x y

x yy

6 12 3 0

6 12 3

6 12 3

12 1212 1

9 9

2

2

2

2

+ - - =

+ = +

+ = +

+ = +

= +

+ +

^ h

So the parabola has equation .x y3 12 12+ = +^ ^h h Its vertex has coordinates , .3 1- -^ h

4 12

3

a

a`

=

=

The parabola is concave upwards as it is in the form .x h a y k4- = -2^ ^h h

Count up 3 units to the focus ,3 2focus` = -^ h Count down 3 units to the directrix ̀ directrix has equation .y 4= -

It is easy to fi nd the focus and the

directrix by counting along the y -axis.

The concave downwards parabola with vertex ( h , k ) and focal length a has equation x h a y k4- = - -2^ ^h h

Proof

Find the equation of the concave downwards parabola with vertex ( h , k ) and focal length a.

ch10.indd 519 7/18/09 4:03:52 PM


Counting down a units from the vertex V gives the focus , .F h k a= -^ h Counting up a units from the vertex V gives the point on the directrix , .D h k a= +^ h So the equation of the directrix is given by .y k a= + We fi nd the equation of the locus of P ( x , y ) that is equidistant from point ,F h k a-^ h and line .y k a= +

y

x

F (h , k- a)

P (x, y)

B y= k+a

B has coordinates , .x k a+^ h

( )

PF PB

PF PB

x h y k a x x y k ax h y k a y k a

x h y k a y k ay k a y k a y k a y k a

y k aay ak

a y k

2 2 24 4

4

We want


2 2

2 2

=

=

- + - - = - + - +

- + - + = - -

- = - - - - +

= - - + - + - - - - +

= - -

= - +

= - -

2 2

2 2 2

2 2 2

^ ^ ^ ^^ ^ ^

^ ^ ^^ ^ ^ ^

^ ^

^

h h h hh h h

h h hh h h h

h h

h

7 7

7 7

A A

A A

ch10.indd 520 7/18/09 4:03:53 PM


The parabola x h a y k4- = - -2^ ^h h has • axis parallel to the y -axis vertex• at ( h , k ) focus• at ,h k a-^ h • directrix with equation y k a= +

EXAMPLES

1. Find the equation of the parabola with focus ( 2- , 1) and directrix .y 3=

Solution

x(-2, 1)

y=3

-2 -1

1

1

B

2

1

y

3

Coordinates of B are ( 2- , 3) . The vertex is the midpoint of ( 2- , 1) and ( 2- , 3). ̀ vertex = ( 2- , 2) Focal length .a 1= From the diagram the curve is concave downwards. The equation is in the form

.

x h a y k

x yx y

x x y

x x y

4

2 4 1 22 4 2

4 4 4 8

4 4 4 0

i.e. 2

2

2

- = - -

- - = - -

+ = - -

+ + = - +

+ + - =

2

2

^ ^^ ] ^^ ^

h hh g hh h

7 A

2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola .x x y8 8 16 02 - + - =

CONTINUED

ch10.indd 521 7/18/09 4:03:54 PM


PARABOLA ( y - k)2 = 4a(x - h)

Solution

Complete the square on x.

x x y

x x y

x x y

x yy

8 8 16 0

8 8 16

8 8 16

4 8 328 4

16 16

2

2

2

2

- + - =

- = - +

- = - +

- = - +

= - -

+ +

^^

hh

So the parabola has equation .x y4 8 4- = - -2^ ^h h Its vertex has coordinates (4, 4).

4 8

2

a

a`

=

=

The parabola is concave downwards as it is in the form .x h a y k4- = - -2^ ^h h

(4, 4)

(4, 2)

1

1

2

3

4

5

2 3 4

2

2

y

y=6

Count down 2 units to the focus ,4 2focus` = ^ h Count up 2 units to the directrix ̀ directrix has equation .y 6=

The parabola with vertex ( h , k ) and focal length a that turns to the right has equation y k a x h4- = -2^ ^h h

Proof

Find the equation of the parabola that turns to the right with vertex ( h , k ) and focal length a.

ch10.indd 522 7/18/09 4:03:55 PM


Counting a units to the right from the vertex V gives the focus , .F h a k= +^ h Counting a units to the left from the vertex V gives the point on the directrix , .D h a k= -^ h So the equation of the directrix is given by .x h a= - We fi nd the equation of the locus of P ( x , y ) that is equidistant from point ,F h a k+^ h and line .x h a= -

y

F (h+a, k)

x

P (x, y)

x=h-a

B

B has coordinates , .h a y-^ h

( )

PF PB

PF PB

x h a y k x h a y yx h a y k x h a

y k x h a x h ax h a x h a x h a x h a

x h aax ah

a x h

2 2 24 4

4

We want


2 2

2 2 2

2

=

=

- + + - = - - + -

- - + - = - -

- = - + - - -

= - + + - - - + - - -

= -

= -

= -

2

2 2

2 2 2

^ ^ ^ ^^ ^ ^

^ ^ ]^ ^ ^ ^

^ ^

^

h h h hh h h

h h gh h h h

h h

h

7 7

7 7

A A

A A

ch10.indd 523 7/18/09 4:03:55 PM


The parabola y k a x h4- = -2^ ^h h has • axis parallel to the x -axis • vertex at ,h k^ h focus• at ,h a k+^ h directrix• with equation x h a= -

EXAMPLES

1. Find the equation of the parabola with focus (1, 1- ) and directrix .x 5= -

Solution

x

(1, -1)

4

2

3

-4 -3 -2-1 1 2 3

33

B4 5

5x=-5

1

y

-5

-2

-3

Coordinates of B are ( 5- , 1- ). The vertex is the midpoint of ( 5- , 1- ) and (1, 1- ). ,2 1vertex` = - -^ h Focal length 3a = From the diagram the parabola curves to the right. The equation is in the form

1 2

y k a x h

y xy x

y y x

y y x

4

4 31 12 2

2 1 12 24

2 12 23 0

i.e.

2

2

2

2

- = -

- - = - -

+ = +

+ + = +

+ - - =

2

^ ]^ ] ^^ ]

h gh g hh g

7 7A A

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2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola .y y x12 4 8 02 + - - =

Solution

Complete the square on y.

y y x

y y x

y y x

y xx

12 4 8 0

12 4 8

12 4 8

6 4 444 11

36 36

2

2

2

+ - - =

+ = +

+ = +

+ = +

= +

+ +2^

^h

h

So the parabola has equation 11y x6 42+ = +^ ^h h .y x6 4 11or 2

- - = - -] ]g g7 6A @ Its vertex has coordinates ( 11- , 6- ).

4 4

1

a

a`

=

=

The parabola turns to the right as it is in the form .y k a x h4- = -2^ ^h h

y

x

(-10,-6)(-11,-6)

x=-12

11

Count 1 unit to the right for the focus , .10 6focus` = - -^ h Count 1 unit to the left for the directrix ̀ directrix has equation .x 12= -

PARABOLA (y – k)2 = – 4a(x – h)

The parabola with vertex ( h , k ) and focal length a that turns to the left has equation y k a x h4- = - -2^ ^h h

ch10.indd 525 7/18/09 4:03:57 PM


Proof

Find the equation of the parabola that turns to the left with vertex ( h , k ) and focal length a.

Counting a units to the left from the vertex V gives the focus ,F h a k= -^ h . Counting a units to the right from the vertex V gives the point on the directrix ,D h a k= +^ h . So the equation of the directrix is given by x h a= + . We fi nd the equation of the locus of P ( x , y ) that is equidistant from point ,F h a k-^ h and line x h a= + .

x=h+a

B

F (h-a, k)

P (x, y)

y

x

B has coordinates , .h a y+^ h

PF PB

PF PB

x h a y k x h a y yx h a y k x h a

y k x h a x h ax h a x h a x h a x h a

We want2 2

=

=

- + - = - + -

- + - = -

- = - - -

= - + - - - -

- +

+ -

- +

- + - +

2 2 2 2

2 2 2

2 2 2

^ ^ ^ ^^ ^ ^

^ ^ ^^ ^ ^ ^

h h h hh h h

h h hh h h h

7 7

7 7

A A

A A

(difference of two squares)

2 2 2x h aax ah

a x h

a y k

4 4

4

4

= - -

= - +

= - -

= - -

^ ^

^^

h h

hh

ch10.indd 526 7/18/09 4:03:58 PM


The parabola y k a x h4- = - -2^ ]h g has • axis parallel to the x -axis • vertex at ( h , k ) focus• at ,h a k-^ h directrix• with equation x h a= +

EXAMPLES

1. Find the equation of the parabola with focus (2, 1) and directrix 3x = .

Solution

x

y

B

x=3

1

1 2

(2, 1)

12

12

12(2 , 1)

Coordinates of B are (3, 1). The vertex is the midpoint of (3, 1) and (2, 1).

,221 1vertex` = c m

Focal length 21a =

From the diagram the parabola curves to the left. The equation is in the form

4

1

2

y a x h

y x

y x

y y x

y y x

k

421 2

21

1 221

2 1 2 5

2 2 4 0

i.e.

2

2

- = - -

- = - -

- = - -

- + = - +

- + - =

2

2

2

^ ^^ c c^ c

h hh m mh m

2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola .y y x4 8 4 02 + + - =

CONTINUED

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Solution

Complete the square on y.

y y x

y y x

y y x

y xx

4 8 4 0

4 8 4

4 8 4

2 8 88 1

4 4

2

2

2

2

+ + - =

+ = - +

+ = - +

+ = - +

= - -

+ +

^]

hg

So the parabola has equation y x2 8 12+ = - -^ ]h g .y x2 8 1or 2

- - = - -] ]g g7 A Its vertex has coordinates , .1 2-^ h

4 8

2

a

a`

=

=

The parabola turns to the left as it is in the form y k a x h4- = - -2^ ^h h

Count 2 units to the left for the focus , .1 2focus` = - -^ h Count 2 units to the right for the directrix ̀ directrix has equation .x 3=

1. Complete the square on x to write each equation in the form .x h a y k42

!- = -] ^g h (a) x x y6 8 15 02 - - - = (b) x x y10 4 1 02 - - + = (c) x x y2 4 11 02 - - - = (d) x x y8 12 20 02 - + - =

(e) x x y12 8 20 02 =- - - (f) x x y14 16 1 02 + + + = (g) x x y4 4 16 02 - + - = (h) x x y18 12 9 02 + - + = (i) x x y2 8 7 02 + - - = (j) x x y6 4 1 02 - + + =

10.6 Exercises

x=3

y

x

(1, -2)

1

-1-2-3 -1

(-1, -2)

2 2

-222222

21 3

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2. Complete the square on y to write each equation in the form y k a x h42

!- = -^ ]h g (a) y y x8 4 02 =- - (b) y y x2 8 15 02 =- - - (c) y y x4 12 8 02 + =- - (d) y y x20 4 16 02 + =- - (e) y y x6 16 7 02 + + =- (f) y y x12 8 4 02 + =- - (g) y y x10 24 23 02 + + =- (h) y y x24 4 02 + =- (i) y y x4 20 16 02 + =- - (j) y y x8 8 02 + + =

3. Find the equation of each parabola focus (a) , ,1 3-^ h directrix 1y = - focus (b) 4, 1 ,-^ h directrix y 1= - focus (2, 0), directrix (c) 4y = - focus (3, 6), directrix (d) 2y = focus (e) 2, 5 ,-^ h directrix

3y = - focus (f) , ,1 4- -^ h directrix 4y = focus (g) 3),( ,4 - directrix 7y = focus (h) 5, 1 ,-^ h directrix 5y = focus (i) , ,3 6- -^ h directrix 0y = focus (j) , ,0 7-^ h directrix 5y = - focus (2, 3), directrix (k) 4x = - focus (l) , ,1 4-^ h directrix 3x = - focus (6, 0), directrix (m) 2x = focus (n) 2( ),,3 - directrix

5x = - focus (o) , ,1 1-^ h directrix 3x = - focus (p) , ,2 4- -^ h directrix 4x = focus (2, 1), directrix (q) 4x = focus (r) 5, 3 ,-^ h directrix 3x = focus (s) 1, 2 ,-^ h directrix 0x = focus (3, 1), directrix (t) x 4=

4. Find (i) the coordinates of the focus and

(ii) the equation of the directrix of (a) x x y6 4 3 02 =- - - (b) x x y2 8 7 02 =- - - (c) x x y4 4 02 + =- (d) x x y8 12 4 02 + =- - (e) x x y10 8 1 02 + + =- (f) x x y6 4 1 02 + + =-

(g) x x y2 8 15 02 + + =- (h) x x y4 4 02 + =- (i) x x y8 12 4 02 + + =- (j) x x y4 16 12 02 + + =-

5. Find (i) the coordinates of the focus and

(ii) the equation of the directrix of (a) y y x2 4 3 02 + =- - (b) y y x8 12 4 02 + =- - (c) y y x6 8 7 02 =- - - (d) y y x4 16 12 02 + =- - (e) y y x2 24 25 02 + =- - (f) y y x10 8 1 02 + + + = (g) y y x14 4 1 02 + + + = (h) y y x12 20 4 02 - + - = (i) y y x4 32 28 02 + =- - (j) y y x6 40 29 02 + + + =

6. Find the equation of the parabola with vertex ,0 3^ h if it is concave upwards and 3.a =

7. Find the equation of the parabola with vertex , ,2 1- -^ h focal length 2, and axis parallel to the y -axis.

8. A parabola has its vertex at ,1 5-^ h and its focal length as 1. If the parabola is concave upwards, fi nd its equation.

9. A parabola has its axis parallel to the x -axis. If its vertex has coordinates ,2 6^ h and ,a 3= fi nd its equation if it turns to the left.

10. Find the equation of the parabola with vertex at ,1 0^ h and focus at , .1 4^ h

11. Find the equation of the parabola that has vertex ,1 1^ h and focus , .1 8^ h

12. A parabola has its vertex at ,2 2-^ h and focus at , .4 2- -^ h Find its equation.

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13. Find the equation of the parabola with vertex ,0 3^ h and focus , .8 3^ h

14. Find the equation of the parabola with vertex ,3 3^ h and equation of directrix .y 5=

15. Find the equation of the parabola with vertex ,3 1-^ h and directrix .x 1= -

16. A parabola has directrix y 5= and focus , .3 3-^ h Find its equation.

17. Find the equation of the locus of a point moving so that it is equidistant from the point ,2 2^ h and the line .y 4= -

18. Find the equation of the parabola with focus ,2 1-^ h and directrix .x 10=

19. Find the coordinates of the vertex and focus and the equation of the directrix for the parabola

(a) x x y4 8 12 02 + - + = (b) x x y6 12 33 02 - - + = (c) x x y2 4 5 02 - + + = (d) y y x8 16 64 02 - - + = (e) y y x4 24 4 02 + - + = (f) .y x8 40 02 + + =

20. For the parabola ,x x y2 28 111 02 + + - = fi nd the coordinates of its vertex and focus, and the equations of its directrix and axis. What is its maximum value?

21. The latus rectum of a parabola has endpoints ,2 3-^ h and , .6 3^ h Find two possible equations for the parabola.

22.

Find the equation of the arch (a) above.

Find the coordinates of its (b) focus and the equation of its directrix.

23. (a) Sketch ,y x x2 82= + - showing intercepts and the minimum point.

Find the coordinates of the (b) focus and the equation of the directrix of the parabola.

24. Find the equation of the parabola with vertex ,2 3-^ h that also passes through ,2 1^ h and is concave downwards.

25. A parabolic satellite dish has a diameter of 4 m at a depth of 0.4 m. Find the depth at which its diameter is 3.5 m, correct to 1 decimal place.

DID YOU KNOW?

The word ‘directrix’ is due to the Dutch mathematician Jan De Witt (1629–72). He published a work called Elementa curvarum , in which he defi ned the properties of the parabola, ellipse, circle and hyperbola. These curves are all called conic sections .

ch10.indd 530 7/18/09 4:08:24 PM


De Witt was well known as the ‘Grand Pensionary of Holland’. He took part in the politics and wars of his time, opposing Louis XIV. When the French invaded Holland in 1672, De Witt was seized and killed.

Tangents and Normals

Remember that the gradient of the tangent to a curve is given by the derivative.

The normal to the curve is perpendicular to its tangent at that point. That is, 1m m1 2 = - for perpendicular lines.

EXAMPLES

1. Find the gradient of the tangent to the parabola 8x y2 = at the point , .4 2^ h

Solution

x y

y x

dx

dy x

x

8

8

82

4

2

2

`

=

=

=

=

CONTINUED

ch10.indd 531 7/18/09 4:08:24 PM


, ,dx

dy4 2

44

1

At =

=

^ h

So the gradient of the tangent at ,4 2^ h is 1.

2. Find the equation of the normal to the parabola 4x y2 = at the point , .8 16-^ h

Solution

( , ):

the .

x

dx

dy x

x

dx

dy

m

4

42

2

8 1628

So

At

So gradient of the tangent

2

2

1

--

x y

y

4

4

4

=

=

=

=

=

= -

= -

The normal is perpendicular to the tangent.

m m

m

m

1

4 1

41

So1 2

2

2

`

`

= -

- = -

=

] g

The equation of the normal is given by

( )

[ ( )]

( )

.

y y m x x

y x

x

y x

x y

1641 8

41 8

4 64 8

0 4 72

i.e.

1 1- = -

- = - -

= +

- = +

= - +

1. Find the gradient of the tangent to the parabola 12x y2 = at the point where 2.x =

2. Find the gradient of the tangent to the parabola 3x y2 = - at the point , .6 12-^ h

3. Find the gradient of the normal to the parabola 4x y2 = at the point where 2.x =

4. Find the gradient of the tangent to the parabola 16x y2 = at the point , .4 1^ h

10.7 Exercises

ch10.indd 532 7/18/09 4:08:26 PM


5. Show that the gradient of the tangent to the curve 2x y2 = at any point is its x -coordinate.

6. Find the equation of the tangent to the curve 8x y2 = at the point , .4 2^ h

7. Find the equation of the normal to the curve 4x y2 = at the point where 4.x = -

8. Find the equations of the tangent and normal to the parabola 24x y2 = - at the point , .12 6-^ h

9. Find the equations of the tangent and normal to the parabola 16x y2 = at the point where 4.x =

10. Find the equation of the tangent to the curve 2x y2 = - at the point , .4 8-^ h This tangent meets the directrix at point M . Find the coordinates of M .

11. Find the equation of the normal to the curve 12x y2 = at the point , .6 3^ h This normal meets the parabola again at point P . Find the coordinates of P .

12. The normal of the parabola 18x y2 = at ,6 2-^ h cuts the parabola again at Q . Find the coordinates of Q .

13. Find the equations of the normals to the curve 8x y2 = - at the

points ,16 32- -^ h and , .221

- -c m

Find their point of intersection and show that this point lies on the parabola.

14. Find the equation of the tangent at ,8 4^ h on the parabola 16 .x y2 = This tangent meets the tangent at the vertex of the parabola at point R . Find the coordinates of R .

15. (a) Show that the point ,P p p2 2_ i lies on the parabola 4 .x y2 =

Find the equation of the (b) normal to the parabola at P .

Show that (c) p 1 02 + = if the normal passes through the focus of the parabola .p 0!^ h

ch10.indd 533 7/18/09 4:08:26 PM


Test Yourself 10 1. Find the equation of the locus of a point

moving so that it is equidistant from ,A 1 2-^ h and ,B 3 5^ h .

2. Find the equation of the parabola with focus ,2 1^ h and directrix .y 3= -

3. Find the radius and centre of the circle 6 2 6 0.x x y y2 2- + - - =

4. Find the coordinates of the vertex and (a) the focus of the parabola (b)

( ) .y x3 12 12+ = -] g

5. Find the equation of the locus of a point that is always 5 units from the origin.

6. Find the equation of the directrix and(a) the coordinates of the focus of the (b)

parabola .x y82 = -

7. A point ,P x y^ h moves so that AP and BP are perpendicular, given ,A 3 2= ^ h and , .B 4 1= -^ h Find the equation of the locus of P .

8. Point ,x yP ^ h is equidistant from the point ,A 4 2-^ h and the line 6.y = Find the equation of the locus.

9. Find (a) the coordinates of the (i) vertex and (ii) focus and (b) the equation of the directrix of the parabola .x x y2 4 5 02 - - + =

10. Find the equation of the tangent to the parabola 18x y2 = at the point , .6 2-^ h

11. Find the length of the diameter of the circle 8 12 3 0.x x y y2 2+ + - + =

12. Find the equation of the parabola with directrix x 6= and focus , .6 0-^ h

13. A parabola has a focus at ,0 4^ h and its vertex is at ,0 2^ h . Find the equation of the parabola.

14. Find the equation of the locus of a point that is always 3 units from the line .x y4 3 1 0- - =

15. A point is equidistant from the x - and y -axis. Find the equation of its locus.

16. Find the equation of the parabola with vertex at the origin, axis 0y = and

passing through the point , .141 5c m

17. Find the gradient of the tangent and (a) the normal to the parabola (b) 12x y2 = -

at the point where 3.x =

18. (a) Find the equation of the normal to the parabola 4x y2 = at the point , .8 16-^ h

This normal cuts the parabola again (b) at Q . Find the coordinates of Q .

19. Show that 7 3 12 0x y- + = is a focal chord of the parabola 16 .x y2 =

20. Find the point of intersection of the normals to the parabola x y122 = - at the

points ,4 131

-c m and , .231

- -c m

21. (a) Find the equation of the tangent to the parabola x y122 = at the point P (6, 3).

Find (b) R , the y- intercept of the tangent. Show that (c) FP FR= where F is the

focus.

ch10.indd 534 7/18/09 4:08:26 PM


1. (a) Find the equation of the locus of point P , which is equidistant from fi xed points ,A 3 5^ h and , .B 1 2-^ h

Show that this locus is the (b) perpendicular bisector of line AB .

2. (a) Find the equation of the circle with centre ,1 3^ h and radius 5 units.

Show that the circle cuts the (b) x -axis at the points ,5 0^ h and , .3 0-^ h

3. The line with equation x y5 12 36 0- + = is a chord of the parabola .x y122 = Find the point of intersection of the tangents to the parabola from the endpoints of the chord.

4. (a) Find the equation of the normals to the parabola x y82 = at the points

,M 221

-c m and , .N 8 8^ h

Show that these normals are (b) perpendicular.

Find the point of intersection (c) X of the normals.

Find the equation of line (d) MN and show that it is a focal chord.

5. From which point on the parabola x ay42 = does the normal pass through the focus?

6. (a) Find the equation of the tangents to

the parabola x y42 = at the points

,A 141c m and , .B 4 4-^ h

Show that the point of intersection of (b) these tangents lies on the directrix.

7. Find the equation of the parabola with axis parallel to the y -axis and passing through points , ,0 2-^ h ,1 0^ h and , .3 8-^ h

8. Find the equation of the straight line through the centres of the circles with equations x x y y4 8 5 02 2+ + - - = and .x x y y2 10 10 02 2- + + + =

9. Sketch the region 2 4 4 0.x x y y2 2 #+ + - -

10. (a) Find the equation of the locus of a point P moving so that PA is perpendicular to PB where ,A 4 3-= ^ h and , .B 0 7= ^ h

Show that this locus is a circle with (b) centre ,2 5-^ h and radius .2 2

11. Find the exact gradient, with rational denominator, of the normal to the parabola y x122 = at the point where x 4= in the fi rst quadrant.

12. (a) Find the equation of the parabola with vertex ,3 2-^ h and focus , .7 2-^ h

Find the equation of the tangent to (b) the parabola at the point where x 4= in the fi rst quadrant.

13. Find the exact length of the line from ,2 7^ h to the centre of the circle .x x y y4 6 3 02 2+ + - - =

14. Find the equation of the locus of midpoints of all chords of length 2 units in the circle with equation .x y y2 3 02 2+ - - =

15. A satellite dish is to be 3.5 m wide and 1.1 m deep. Find the position of the focus in millimetres, correct to the nearest millimetre.

3.5 m

1.1 m

16. Find the equation of the locus of point P that moves such that the distance from P to the lines 3 4 1 0x y- + = and 12 5 3 0x y+ + = is in the ratio : .3 1

Challenge Exercise 10

ch10.indd 535 7/20/09 10:38:50 AM

maths in focus - margaret grove - ch10

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