Simultaneous linear and quadratic equations and loci
32 CHAPTER
32.1 Solving simultaneous equations Section 10.4 covered solving simultaneous equations where both equations are linear, that is, finding a pair of values of x and y which are solutions of both equations. This section covers solving a pair of simultaneous equations where one is linear and one is quadratic. Consider the equations y 2x 1 and y x 2 4
y 2x 1 is the linear equation, because it has the form y mx c and y x 2 4 is the quadratic equation, because it has the form y ax2 bx c To solve these simultaneous equations substitute 2x 1 for y in y x 2 4 This gives 2x 1 x 2 4 This quadratic equation in x can be solved using one of the methods covered in Chapter 30. Firstly, write 2x 1 x 2 4 in the form ax 2 bx c 0 x 2 2x 3 0 (x 3)(x 1) 0 so x 3 or x 1 To find the values of y, substitute into the linear equation y 2x 1 When x 3, y 5 and when x 1, y 3 Sometimes the linear equation must be rearranged to make y the subject.
Example 1 Solve the simultaneous equations y 2x 2 and y 2x 4
Solution 1 y 4 2x
Make y the subject of y 2x 4
4 2x 2x 2
Substitute 4 2x for y in y 2x2
2x 2 2x 4 0
Rearrange into the form ax2 bx c 0
x2 x 2 0
Divide both sides by 2
(x 2)(x 1) 0
Factorise.
x 2 or x 1 When x 2, y 8
To find the values of y, substitute into y 4 2x
When x 1, y 2 521
Simultaneous linear and quadratic equations and loci
CHAPTER 32
Example 2 Solve y 4x2 and 3y 2x 4
Solution 2 4 2x y 3 4 2x 4x 2 3
4 2x Substitute for y in y 4x2 3
4 2x 12x2
Multiply both sides by 3
12x2 2x 4 0
Rearrange.
6x2 x 2 0
Divide both sides by 2
(3x 2)(2x 1) 0
Factorise.
Make y the subject of 3y 2x 4
x 23 or x 12 2
When x 23, y 4 (23) 196 2
When x 12, y 4 (12) 1
To find the values of y, substitute into y 4x2 as the working 4 2x required is easier than substituting into y 3
Using simultaneous equations to find the intersection of a line and a quadratic curve Just as the solution of two simultaneous linear equations represents the point of intersection of two straight lines, so the solution of a pair of simultaneous equations where one is linear and one is quadratic represents the points of intersection of a straight line and a quadratic curve. The sketch shows the straight line y x 2 and the quadratic curve y x2. The line and the curve intersect at two points. The coordinates of these points of intersection satisfy both equations y x 2 and y x 2 simultaneously (at the same time) and can be found by solving the simultaneous equations y x 2 and y x2.
quadratic curve
y
straight line
6 4 2 4
2
O
2
4 x
x2 x 2 x x20 (x 2)(x 1) 0 x 2 or x 1 2
so
When x 2, y 4 and when x 1, y 1 So the coordinates of the points of intersection of the line and the curve are (2, 4) and (1, 1).
Example 3 Find the coordinates of the points of intersection of the straight line y x and the curve y 3x2 2
Solution 3
x 3x 2 2
3x2 x 2 0 (3x 2)(x 1) 0
x 23 or x 1 522
Solve the simultaneous equations y x and y 3x2 2
32.2 Loci and equations
CHAPTER 32
When x 23, y 23
y
To find the values of y, substitute into y x
10
When x 1, y 1 Points of intersection are (23, 23) and (1, 1)
5
2
Exercise 32A 1 Solve these simultaneous equations. a y 2x and y 2x 2 c y 3x 1 and y x 2 5 2 Solve a y 3x 4 and y 2x 2 5 c y 2x 5 and y 2x 2 x e 2x y 8 and y x 2
1
O
1
2 x
b y x 3 and y x 2 3x d y 6 x 2 and y 4x 1 b x y 2 and y 3x 2 2 d x y 3 and y x2 2x 1 f 2x 3y 13 and y x 2 1
3 Find the coordinates of the points of intersection of these lines and quadratic curves. b y 5 and y x2 4x a y 3 and y x 2 2x c y 4 and y x2 5x d y 1 and y 2x 2 5x 1 4 Find the coordinates of the points of intersection of these lines and quadratic curves. b y x and y x 2 2 a y x 6 and y x 2 2 d y 4 x and y 2x 2 3 c y x 1 and y = 2x f x 2y 0 and y 2x2 4x 1 e y x and y x 2 7x 5
32.2 Loci and equations Locus (plural ‘loci’) was considered in Section 27.2 A locus is the path of a point which moves according to some mathematical rule. Alternatively, it is the set of points which obey a mathematical rule. The following examples show how to find the equations of some loci.
Example 4 Find the equation of the locus of a point which moves so that it is equidistant from the points A(2, 2) and B(8, 2).
Solution 4 The equation of the locus is x 5
The locus of a point which moves so that it is equidistant from two fixed points A and B is the perpendicular bisector of AB. Draw this perpendicular bisector and write down its equation.
y 4 2
O
A
B
2
4
6
8
10 x
2
523
Simultaneous linear and quadratic equations and loci
CHAPTER 32
Example 5 a Sketch the locus of a point which is equidistant from the points P(4, 0) and Q(0, 4). b Find the equation of this locus.
Solution 5 y 6
Q
4 2
P O
2
6 x
4
The equation of the locus is y x
The locus is the line which passes through (0, 0), (1, 1), etc.
Sometimes the equation of a locus is obtained by considering a point with coordinates (x, y) representing any point on the locus.
Example 6 P is the point (x, y).
y
a Write down, in terms of y, an expression for the distance of P from the line y 2
8
b Write down, in terms of x and y, an expression for distance of P from the point (0, 4)
6
The distance of the point P from the line y 2 is the same as its distance from the point (0, 4).
4
d Show that the equation of the locus can be written as
Solution 6 a y
y2
2
c Write down an equation which describes this locus.
x2 12 y 4
P(x, y)
O
2
P(x, y)
y2 y y2 2
O
x
Distance of P from the line y 2 is y 2 524
The perpendicular distance is the difference between the y values.
4
6 x
32.2 Loci and equations
CHAPTER 32
2 b Distance of P from the point (0, 4) x ( y 4)2
Use Pythagoras’ theorem.
y
P(x, y) y4
4
x
x
O 2 c x ( y 4)2 y 2
d
The distances in a and b are the same.
x2 ( y 4)2 ( y 2)2
Square both sides of the equation in part c.
x2 y2 8y 16 y2 4y 4
Expand the brackets.
x 2 12 4y
Simplify.
x2 12 y 4
Divide both sides by 4
y x2 12 x2 may be written as y 3 4 4
The locus of a point P which moves so that its distance from a fixed point O is always 4 cm is a circle, centre O and radius 4 cm. y
To find the equation of a circle, centre O(0, 0) and radius 4, let P(x, y) be any point on the circle. Then OP 4
4
P(x, y)
Using Pythagoras’ theorem x y 4 2
2
2
So the equation of the circle is x2 y2 16
4
4
O
y x
4
x
4
In general, the equation of a circle, centre O and radius r, is x2 y2 r 2 In Chapter 23, the x-coordinates of the points of intersection of a line and a quadratic curve were used to estimate the solutions of the related quadratic equation. In a similar way, the coordinates of the points of intersection of a straight line and a circle, centre the origin, can be used to find estimates of the solutions to the pair of simultaneous equations representing the line and the circle. 525
Simultaneous linear and quadratic equations and loci
CHAPTER 32
Example 7 a Draw the circle with equation x 2 y 2 25 b On the same axes, draw the line with equation y 2x 1 c Hence find estimates of the solutions to the simultaneous equations x2 y2 25 y 2x 1 Give your answers correct to 1 decimal place.
Solution 7 a
The circle has centre O and radius 5
y 6 4 2
6
4
2
O
2
4
6 x
2 4 6
b
y
y 2x 1
Draw the straight line with equation y 2x 1
6 4 2
6
4
2
O
2
4
6 x
2 4 6
c Estimates of the solutions to the simultaneous equations are x 1.8, y 4.6 or x 2.6, y 4.2
The estimates are obtained from the coordinates of the points of intersection of the straight line and the circle.
Exercise 32B 1 In each case sketch the locus of the point which moves so that it is equidistant from the given points. Give the equation of each locus. a (0, 0) and (6, 0) b (2, 0) and (6, 0) c (0, 0) and (0, 4) d (0, 4) and (0, 2) 526
32.3 Intersection of lines and circles – algebraic solutions
CHAPTER 32
2 In each case sketch the locus of the point which moves so that it is equidistant from the given points. Give the equation of each locus. a (3, 0) and (0, 3) b (4, 0) and (0, 4) c (2, 3) and (3, 2) d (1, 4) and (4, 1) 3 Write down the equation of the circle, centre (0, 0), radius 6 4 P is the point (x, y) a Write down the distance of P from the y-axis. b Write down the distance of P from the x-axis. c Find the equation of the locus of P for each of these rules. i The point P moves so that its distance from the y-axis is the same as its distance from the x-axis. ii The point P moves so that its distance from the y-axis is twice its distance from the x-axis. iii The point P moves so that its distance from the y-axis is 1 more than its distance from the x-axis. 5 P is the point (x, y). a Write down an expression for the distance of P from the line y 1 b Write down an expression for the distance of P from the point (0, 3).
P moves so that its distance from the line y 1 is the same as its distance from the point (0, 3). c Find the equation of the locus of P. Give your answer in its simplest form. 6 The point P(x, y) moves so that its distance from the line x 1 is the same as its distance from the line y 2. Find the equation of the locus of P. 7 a Draw on graph paper the circle with equation x 2 y2 16 b Using the same axes, draw the straight line with equation y x 1 c Hence find estimates of the solutions to the simultaneous equations x 2 y2 16 and yx1 8 Draw suitable graphs to find estimates of the solutions to these pairs of simultaneous equations. b x 2 y2 64 and 3x 2y 12 a x 2 y2 16 and y 2x 9 By drawing suitable graphs, solve these simultaneous equations. b x 2 y2 100 and x y 2 a x2 y2 25 and y x 1
32.3 Intersection of lines and circles – algebraic solutions The coordinates of the points of intersection of a straight line and a circle can be found by solving the simultaneous equations representing the line and the circle. For example, the solutions of the simultaneous equations x 2 y2 17 and y x 3 represent the coordinates of the points of intersection of the circle and the line. Substitute x 3 for y in x2 y2 17
x2 (x 3)2 17 x x2 6x 9 17 2x2 6x 8 0 x2 3x 4 0 (x 4)(x 1) 0 so x 4 or x 1 2
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Simultaneous linear and quadratic equations and loci
CHAPTER 32
Substitute into the linear equation y x 3 to obtain the corresponding values of y.
6
When x 4, y 1 When x 1, y 4
4
y yx3
So the line and circle intersect at (4, 1) and (1, 4). 2
6
4
2
O
2
4
6 x
2 4
x2 y2 17
6
Example 8 Solve x2 y2 20 and x 2y 5 Give your answers correct to 3 significant figures.
Solution 8 x 5 2y (5 2y)2 y2 20 25 20y 4y2 y2 20 5y2 20y 5 0 y2 4y 1 0
Make x the subject of x 2y 5 Substitute 5 2y for x in x2 y2 20
Simplify.
2 (4) (4) 4 11 y 21 4 12 y 2 y 3.7320… or y 0.2679…
Substitute a 1, b (4), c 1 into 2 c 4a b b the quadratic formula x 2a
When y 3.73, x 2.46
Substitute into x 5 2y
It is equally correct to make y the subject of x 2y 5 and then substitute 5x for y in 2 x2 y2 20 but the algebra is more complicated.
Find y with a calculator.
When y 0.268, x 4.46
Exercise 32C 1 Solve these simultaneous equations. b x2 y2 20 a x2 y2 10 yx4 yx6
c x2 y2 26 y 2x 3
d x2 y2 40 yx8
2 Find algebraically the coordinates of the points of intersection of each circle and straight line. b x2 y2 25 c x2 y2 34 d x2 y2 13 a x2 y2 45 yx3 xy7 x 2y 1 2x y 7 3 Solve these simultaneous equations. Give your answers correct to 3 significant figures. b x 2 y2 30 c x 2 y2 40 d x 2 y2 50 a x 2 y2 25 xy6 x 2y 7 yx2 2x 4y 9 528
Chapter 32 review questions
CHAPTER 32
4 a x 2 y2 36 x7 i By eliminating x, show that these simultaneous equations have no solution. ii On the same axes, sketch the graphs of x 2 y2 36 and x 7 b x 2 y2 36 x6 i By eliminating x, show that these simultaneous equations have only one solution. ii On the same axes, sketch the graphs of x 2 y2 36 and x 6
Chapter summary You should know and be able to use these facts:
that the solutions of a pair of simultaneous equations correspond to the coordinates of the points of intersection of the related graphs
that the equation of a circle, centre the origin, radius r is x2 y2 r2
that a circle is the locus of the set of points which are equidistant from a fixed point (the centre). You should also be able to:
solve algebraically or graphically a pair of simultaneous equations ● when one is linear and one is quadratic ● when one is linear and one is of the form x2 y2 r2 (the equation of a circle, centre the origin)
find the equation of a locus from its rule.
Chapter 32 review questions 1 Solve a y 2x and y x 2 c y 2x and y x 2 x
b y 2x and y x 2 3 d y x and y 2x 2 3
2 Solve these simultaneous equations. Give your answers correct to 3 significant figures. b y 2x and y 2x 2 1 a y 2x and y x 2 2 2 d y 2x and y 4 x 2 c y 2x and y x x 1 3 Find the coordinates of the points of intersection of each straight line and curve. b y 3x 2 and y x 2 a y 2x and y = 2x(x 1) 2 c y 3x 3 and y 4x x d x y 2 and y 4x 2 3 4 Solve y 12x2 7 5x y 10 5 Write down the equation of the circle, centre the origin and diameter 12 6 Find the equation of the locus of the point P which moves so that its distance from the point (5, 0) is the same as its distance from the point (0, 5). 7 Find the equation of the locus of the point P which moves such that its distance from the x-axis is the same as its distance from the line y 3 529
Simultaneous linear and quadratic equations and loci
CHAPTER 32
8 Find the equation of the locus of the point P which moves such that its distance from the x-axis is the same as its distance from the line x 3 9 Find the equation of the locus of the point P which moves such that its distance from the x-axis is three times its distance from the y-axis. 10 P is the point (x, y). a Write down an expression for the distance of P from the line y 6 b Write down an expression for the distance of P from the origin. The point P moves such that its distance from the line y 6 is the same as its distance from the origin. c Find the equation of the locus of P. Give your answer in its simplest form. 11 a Draw the circle with radius 6 and centre the origin. b On the same axes, draw the straight line with equation y 2x 1 c Use the graphs to find estimates of the solutions to the simultaneous equations x 2 y2 36 and y 2x 1 12 Draw suitable graphs to find estimates of the solutions to these simultaneous equations. b x 2 y2 49 c x 2 y2 64 a x 2 y2 25 yx2 y 2x 2 x y 10 13 Solve these simultaneous equations. b x 2 y2 80 a x 2 y2 45 yx3 yx4
c x 2 y2 58 y 2x 1
14 The diagram shows a sketch of a curve. The point P(x, y) lies on the curve. The locus of P has the following property: the distance of the point P from the point (0, 2) is the same as the distance of the point P from the x-axis. Show that y 14x2 1
y 4
P(x, y) 2
4
2
O
2
4 x
(1387 June 2004)
15 The line y 4 4x intersects the curve y 3(x x) at the points A and B. Use an algebraic method to find the coordinates of A and B. 2
(1387 November 2005)
16 Use the resource sheet which shows a grid where 4 x 4 and 4 y 4 a On the grid, draw the curve with equation x 2 y2 9 b By drawing a suitable straight line find estimates for the solutions of the equations x 2 y2 9 yx1 17 Bill said that the line y 6 cuts the curve x2 y2 25 at two points. a By eliminating y show that Bill is incorrect. b By eliminating y find the solutions of the simultaneous equations x 2 y2 25 y 2x 2
(1387 June 2004)
18 Solve the simultaneous equations
x 2 y2 29 yx3 530
(1387 November 2003)
