Intersection Of Lines And Parabolas(Tangents To Curves)

To find the point(s) of intersection between a parabola and a straight line , put the two equations together and rearrange to form a quadratic equation with the right -hand side = 0 .

: LineStraight of Eqn.

: Parabola of Eqn. 2

Kmxy

CBxAxy

So, point(s) of Intersection given by

2 KmxCBxAx

02 kCmxBxAx

0)(2 kCxmBAx

Solving this equation will produce any point(s) of intersection.

We can also make use of previous results using the DISCRIMINANT

There are two real and distinct roots :-

So, two points of intersection.

There is only one real root, so only one point of intersection..

So , line is a tangent.

CONDITION RESULT

There are no real roots.So there are no points of intersection.

04 If 2 acb(a)

04 If 2 acb(b)

04 If 2 acb(c)

Diagrams for each of these situations would be as follows :-

(a) two points of intersection.

(b) one point of intersection.(Tangent)

(c) No point of intersection

Example 1

contact. ofpoint thefind and 6

parabola theo tangent ta is 715 that Show2

xxy

xy

SOLUTION

xy

xxy

715

62 Point of intersection when

6 715 2 xxx

0962 xx

9

6

1

c

b

aacb 42 = (-6)2-(4 x 1 x 9)

= 36 - 36

= 0

Since b2- 4ac = 0 there is only one point of intersection , sothe line is a tangent .

To find the point of contact solve the quadratic equation thatrepresents the intersection between the line and the parabola.

0962 xxIn this example this is

(x - 3)(x - 3) = 0

x = 3

Substitute this x-value into either the parabola equation or the line equation to find the y-coord. of the point of contact.

y = 15 - 7 x 3= 15 - 21

= - 6The point of contact is ( 3 , -6)

Example 2 The line y = -2x + k is a tangent to the parabola y = 4x - x2 . Find the value of k.

Solution y = -2x + k meets y = 4x - x2 where

-2x + k = 4x - x2

x2 - 6x + k = 0

Since the line is a tangent b2- 4ac = 0

kc

b

a

6

1 (-6)2 - (4 x 1 x k) = 0

36 - 4k = 0

4k = 36

k = 9

So, the equation of the tangent is y = -2x + 9 OR y = 9 - 2x