S.4 Mathematics

x + y –7 = 0

2x – 3y +6=0

x

y

0

(3, 4)

Put (3,4) into x +y –7 =0

LHS = 3+4 – 7

Put (3,4) into 2x –3y +6 =0

LHS = (2)3 – 3(4) + 6

= 0

(3,4) is the solution of the equations of x +y –7 =0 and 2x –3y +6 =0

= 0

= RHS= RHS

x + y –7 = 0

2x – 3y + 6 = 0

x

y

0

(3, 4)

What is the solution of the simultaneous equations?

0632

07

yx

yx

x

y

0

Two points of intersection

x

y

0

One point of intersection

x

y

0

What is the relationship between the number of points of intersection and the value of discriminant?

No points of intersection

x

y

0

x

y

0 y

0

Case 1:2 points of intersection

∆ > 0

Case 2: 1 point of intersection

Case 3:No point of intersection

∆ = 0

∆ < 0

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

842 xxy

52 xy

Example

52

842

xy

xxy

52 84 2 xxx

052842 xxx

)3(4)6( nt,Discrimina 2 0241236

There are two points of intersection

No need to solve the eq.

0362 xx

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

322 xxy

63 xy

63

322

xy

xxy

63 32 2 xxx

063322 xxx

)3(41 nt,Discrimina 2 011121

There is no point of intersection

032 xx

# 1

42

22

xy

xxy

42 2 2 xxx

04222 xxx

)6)(1(4)1( 2 025241

There are two points of intersection

062 xx

# 2

0122

822

yx

xxy

122 82 2 xxx

0122822 xxx

)4(4)4( 2 01616

There is one point of intersection

0442 xx

# 3

x 0 – 1 – 2

y – 6 – 3 0

y = – 3 x – 6

No point of intersection

x 0

y 0

y = 2 x – 4

Two points of intersection

– 4

2 3

2

x

y

2 x – y – 12 = 0

One point of intersection

– 8

3 4 2

– 6 – 4

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

22 xxy

42 xy

I) Graphical method

II) Discriminant method

We can use :

Any other method?

Determine the number of points of intersection of the parabola and the straight line.Parabola:

Straight line:

22 xxy

42 xy

I) Graphical method

II) Discriminant method

We can use :

Any other method?

42

22

xy

xxy

42 2 2 xxx04222 xxx062 xx

#2

062 xx0)2)(3( xx

23 xorx

∴(3,4) and (-2,-8) are the 2 points of intersection.

84 yory

Which method is the fastest in determining the number of points of intersection of the parabola and the straight line?

I) Graphical method

II) Discriminant method

III) Solving the simultaneous equations (Algebraic method)

1. If the parabola y = – x2 + 2x + 5 and the line y = k intersect at one point, find the value of k.

Exercise

ky

xxy 522

kxx 52 2

0522 kxx)5)(1(4)2( 2 k

kk 424)5(44

24 – 4k = 0

# Ex.1

If the parabola and the line intersect at one point , then the discriminant equals to zero.

∴ k = 6

2. If the straight line y = 3x + k does not cut the parabola y = x2 – 3 x + 2 at any point, find the range of values of k.

Exercise

There is no point of intersection

kxy

xxy

3

232

kxxx 3 23 2

0262 kxx)2(4)6( 2 k

kk 4284836

28 + 4k < 0

# Ex.2

There is no point of intersection so the discriminant is less than zero.

∴ k < – 7

3. If the straight line 2x – y – 1 = 0 cuts the parabola y = 3 x2 + 5x + k at two points, where k is an integer. Find the largest value of k.

Exercise

12

53 2

xy

kxxy

12 5 3 2 xkxx0133 2 kxx)1(1232 k

kk 12312129

– 3 – 12 k > 0

# Ex.3

There are two points of intersection so the discriminant is greater than zero.

∴ k < – 0.25The largest value of k is – 1