Transcript S5Quadratics2

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 .
Eqn. of Parabola
: y  Ax 2  Bx  C
Eqn. of Straight Line
: y  mx  K
So, point(s) of Intersection given by
Ax 2  Bx  C  mx  K
Ax2  Bx  mx  C  k  0
Ax 2  ( B  m ) x  C  k  0
Solving this equation will produce any point(s) of intersection.
We can also make use of previous results
using the DISCRIMINANT
CONDITION
(a)
If b2  4ac  0
RESULT
There are two real and distinct roots :So, two points of intersection.
(b)
If b2  4ac  0
There is only one real root, so only
one point of intersection..
So , line is a tangent.
(c)
If b 2  4ac  0
There are no real roots.
So there are no points of intersection.
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
Show that y  15  7 x is a tangent t o the parabola
y   x 2  x  6 and find the point of contact.
SOLUTION
y   x 2  x  6

y  15  7 x

Point of intersection when
15  7 x   x 2  x  6
x2  6x  9  0
a 1 

b  6
c  9 
b 2  4ac = (-6)2-(4 x 1 x 9)
= 36 - 36
=0
Since b2- 4ac = 0 there is only one point of intersection , so
the line is a tangent .
To find the point of contact solve the quadratic equation that
represents the intersection between the line and the parabola.
In this example this is
x2  6x  9  0
(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
= -6
The point of contact is ( 3 , -6)
Example 2
Solution
The line y = -2x + k is a tangent to the parabola
y = 4x - x2 . Find the value of k.
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
a 1 

b  6
c  k 
(-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