The document discusses determining tangency between curves and lines using the discriminant. It shows: 1) How to prove a line is tangent to a curve by setting their equations equal and requiring the discriminant of the resulting quadratic equation to be 0. 2) Examples that find the condition for tangency (the value of t) and the point of contact between a line y=mx+t and parabolas. 3) The process involves setting the equations equal, factorizing the quadratic, and setting the discriminant equal to 0 to determine t and the point where the factors are 0.

1. Block 2
Tangents and the Discriminant

2. What is to be learned?
• How to prove tangency
• Find conditions for tangency
• Find points of contact for tangents

3. Tangents

4. Tangents
Meets at one point only
 One solution
Using Discriminant
b2
– 4ac = 0

5. Show that y = 6x – 9 is tangent to
curve y = x2
and find point of contact

6. Show that y = 6x – 9 is tangent to
curve y = x2
and find point of contact
y = y
= x2
x2
– 6x + 9 = 0
c.f ax2
+ bx + c = 0
a = 1, b = -6, c = 9
for tangency
(-6)2
– 4(1)(9)
= 0 as required
Discriminant
Must be in form a
+ bx + c = 0
b2
– 4ac = 0
6x – 96x – 9 – +0

7. Show that y = 6x – 9 is tangent to
curve y = x2
and find point of contact?
y = y
6x – 9 = x2
x2
– 6x + 9 = 0
c.f ax2
+ bx + c = 0
a = 1, b = -6, c = 9
for tangency
(-6)2
– 4(1)(9)
= 0 as required
Discriminant
Must be in form
ax2
+ bx + c = 0
b2
– 4ac = 0
need x and y

8. Show that y = 6x – 9 is tangent to
curve y = x2
and find point of contact?
y = y
6x – 9 = x2
x2
– 6x + 9 = 0
(x – 3)(x – 3) = 0
x = 3
need x and y
Find x

9. Show that y = 6x – 9 is tangent to
curve y = x2
and find point of contact?
y = y
6x – 9 = x2
x2
– 6x + 9 = 0
(x – 3)(x – 3) = 0
x = 3
need x and y

10. Show that y = 6x – 9 is tangent to
curve y = x2
and find point of contact?
y = y
6x – 9 = x2
x2
– 6x + 9 = 0
(x – 3)(x – 3) = 0
x = 3
y = 6(3) – 9 or y = 32
= 9 = 9need x and y
Find y

11. The Discriminant and Tangency
Tangents meet curves at one point
One solution
For tangency
b2
– 4ac = 0

12. Show that y = 8x – 17 is tangent to curve
y = x2
– 1 and find point of contact
y = y
8x – 17 = x2
– 1
x2
– 8x + 16 = 0
c.f ax2
+ bx + c = 0
a = 1, b = -8, c = 16
for tangency
(-8)2
– 4(1)(16)
= 0 as required
Discriminant
Must be in form
ax2
+ bx + c = 0
b2
– 4ac = 0

13. Point of Contact
Using x2
– 8x + 16 = 0
(x – 4)(x – 4) = 0
x = 4
Using y = x2
– 1
= 42
– 1
= 15
PoC (4 , 15)
y=x2
– 1
y = 8x – 17
(4,15)
x?
y?
(can use either
original equation)

14. Exam Type Stuff

15. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C

16. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C
a)Point of Intersection
y = y
3x + t = x2
+ 4

17. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C
a)Point of Intersection
y = y
3x + t = x2
+ 4
0 = x2
+ 4 – 3x – t
x2
– 3x + 4 – t = 0( ) as required QED

18. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C
a)Point of Intersection
y = y
3x + t = x2
+ 4
0 = x2
+ 4 – 3x – t
x2
– 3x + 4 – t = 0 as required QED( )

19. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C
b) x2
– 3x + (4 –t) = 0
For tangency
c.f. ax2
+ bx + c = 0
a = 1, b = -3,
(-3)2
– 4(1)(4 – t) = 0
9 – 4(4 – t) = 0
9 – 16 + 4t = 0
4t = 7
t = 7
/4
b2
– 4ac= 0
c = 4 – t

20. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C
b) x2
– 3x + (4 –t) = 0
For tangency
c.f. ax2
+ bx + c = 0
a = 1, b = -3,
(-3)2
– 4(1)(4 – t) = 0
9 – 4(4 – t) = 0
9 – 16 + 4t = 0
4t = 7
t = 7
/4
b2
– 4ac= 0
c = 4 – t
t = 7
/4

21. a) Prove that the line y = 3x + t meets the
parabola y = x2
+ 4 where x2
– 3x + (4 – t) = 0
b) Find t, when line is a tangent and P of
C b) x2
– 3x + (4 –t) = 0
x2
– 3x + (4 – 7
/4) = 0
x2
– 3x + 9
/4 = 0
(x – 3
/2)(x – 3
/2)= 0
x = 3
/2
y = x2
+ 4
= (3
/2)2
+ 4
= 9
/4 + 4
= 6 ¼
t = 7
/4
P of C (1½ , 6¼)

22. a) Prove that the line y = 4x + t meets the
parabola y = x2
+ 2 where x2
– 4x + (2 – t) = 0
b) Find t, when line is a tangent and P of
C
a)Point of Intersection
y = y
4x + t = x2
+ 2
0 = x2
+ 2 – 4x – t
x2
– 4x + 2 – t = 0( ) as required

23. a) Prove that the line y = 4x + t meets the
parabola y = x2
+ 2 where x2
– 4x + (2 – t) = 0
b) Find t, when line is a tangent and P of
C
b) x2
– 4x + (2 – t) = 0
For tangency
c.f. ax2
+ bx + c = 0
a = 1, b = -4,
(-4)2
– 4(1)(2 – t) = 0
16 – 4(2 – t) = 0
16 – 8 + 4t = 0
4t = -8
t = -2
b2
– 4ac= 0
c = 2 – t
Find t

24. a) Prove that the line y = 4x + t meets the
parabola y = x2
+ 2 where x2
– 4x + (2 – t) = 0
b) Find t, when line is a tangent and P of
C b) x2
– 4x + (2 – t) = 0
x2
– 4x + (2 + 2) = 0
x2
– 4x + 4 = 0
(x – 2)(x – 2)= 0
x = 2
y = x2
+ 2
= 22
+ 2
= 4 + 2
= 6
t = -2
P of C (2 , 6)
Find PoC