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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