[email protected] wrote:The equation x = 2y2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is
Let the line l be defined as:
y = mx + c
Here slope = m = 3
y = 3x + c ____________ (1)
Since this intersects the parabola at
x = -5
-5 = 2y^2 + 5y - 17
2y^2 + 5y - 12 = 0
y = (-5 +/- sqrt(25 + 96))/4
y = (-5 +/- sqrt(121))/4
y = (-5 +/- 11)/4
y = -4 or y = 3/2
Since the point of intersection lies on the upper left quadrant, y > 0.
So
y = 3/2
Substitute the values of x and y in
3/2 = 3(-5) + c
3/2 = -15 + c
c = 33/2
So the equation of the line is:
[spoiler]y = 3x + (33/2)[/spoiler]
Cheers
