Tangents (and the discriminant)

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

Tangents

Tangents Meets at one point only  One solution Using Discriminant b 2 – 4ac = 0

Show that y = 6x – 9 is tangent to curve y = x 2 and find point of contact

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

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

Show that y = 6x - 9 is tangent to curve y = x 2 and find point of contact? y = y y = y 6x – 9 = x 2 6x – 9 = x 2 x 2 – 6x + 9 = 0 (x – 3)(x – 3) = 0 x = 3 need x and y

Show that y = 6x - 9 is tangent to curve y = x 2 and find point of contact? y = y y = y 6x – 9 = x 2 6x – 9 = x 2 x 2 – 6x + 9 = 0 (x – 3)(x – 3) = 0 x = 3 need x and y

Show that y = 6x - 9 is tangent to curve y = x 2 and find point of contact? y = y y = y 6x – 9 = x 2 6x – 9 = x 2 x 2 – 6x + 9 = 0 (x – 3)(x – 3) = 0 x = 3 y = 6(3) – 9 or y = 3 2 = 9 = 9 = 9 = 9 need x and y

The Discriminant and Tangency Tangents meet curves at one point  One solution For tangency b 2 – 4ac = 0

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

Point of Contact Using x 2 – 8x + 16 = 0 (x – 4)(x – 4) = 0 x = 4 Using y = x 2 – 1 = 4 2 – 1 = 15 PoC (4, 15) y=x y = 8x - 17 (4,15) x?x? y?y?

Exam Type Stuff

a) Prove that the line y = 3x + t meets the parabola y = x where x 2 – 3x + (4 – t) = 0 b) Find t, when line is a tangent and P of C

a)Point of Intersection y = y 3x + t = x 2 + 4

a) Prove that the line y = 3x + t meets the parabola y = x where x 2 – 3x + (4 – t) = 0 b) Find t, when line is a tangent and P of C a)Point of Intersection y = y 3x + t = x = x – 3x – t x 2 – 3x + 4 – t = 0 ( )as required QED

a) Prove that the line y = 3x + t meets the parabola y = x where x 2 – 3x + (4 – t) = 0 b) Find t, when line is a tangent and P of C a)Point of Intersection y = y 3x + t = x = x – 3x – t x 2 – 3x + 4 – t = 0 ( )as required QED

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

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

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

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

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

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