Florida Gator Cake Templates

Zadanie nr8

Dana is a simple y = 2x + m
a). Determine the number of points in common straight line y = 2x + m parabola
y = x2-2x-3 depending on the parameter m.
b). On the line y = 2x + m, where m = 1
selected segment AB where A length of the interval (1, 3). Set point B.
c). Solve the system of equations:
{kx-2y = 3x + 3
ky = 2
For which values \u200b\u200bof the parameter k a simple cross in the second quadrant of the coordinate system


solution to the problem:
a). Number of points in common and simple parabola is equal to the number of solutions
system of equations:
{y = 2x + y = m
x2-2x-3
number of solutions is equal to the number of solutions
equation: 2x + m = x2-2x-3
ie x2-4x-3-m = 0
This equation is quadratic equation and the number of solutions depends on distinguishing

= 16-4 (3-m) = 16 12 4 28 m = 4m
If> 0 <=> 4m 28> 0 <=> m> -7, this equation, and thus
system has two solutions
<=> If = 0 m = 7, then the system has a solution
If <0> m <7, to układ nie ma rozwiązań
Re. For m> 7 parabola and straight are two common points
For m = 7 parabola and straight to a point in common
For m <7 parabola i prosta nie mają punktów wspólnych
b). On the line y = 2x a selected segment AB. Point A = (1, 3),
point B lies on a straight line y = 2x +1 in the distance from point A, ie, AB =

Let B = (x, y) y = 2x +1 (since point B lies on a straight line) so
B = (x, 2x +1)
AB =
=
is: (x-1) 2 + (2x-2) 2 = 45 x 2-2x
a 4 x 2-8x +4 = 45
5x2-10x-40 = 0 /: 5
x2 2x-8 = 0
= 4 32 = 36 => x = (2-6) / 2 = 2 or x = (2 +6) / 2 = 4
Hence: B = (-2, 2ŮÚ (-2 ) 1) or B = (4, 2ŮÚ4 +1)
or
resp. B = (-2, -3) and B = (4, 9)
c). We solve the system of equations:
{kx-2y = 3 / k 0 iMN
3x + ky = 2 / ŮÚ2 (by opposing factors
we determine x) {
k2x-2ky
6x = 3k 2 ky = 4
________________+
x (k2 +6) = 3k
Since k2 4 6 0 for k R,
We set y by opposing factors
{ KX-2y = 3 / uu (-3)
3x + ky = 2 / k 0 iMN
For k = 0
{0x-2y = 3
0 y = 3x 2
{x = 2 / 3 y =-
3 / 2 This solution does not satisfy the condition
point belonging to the second quarter. Solution


and to appoint a point from the second quarter
, ie x 0, ie