Parametrics...?

Question

Given the parabola x² = (1/2)y, for what values of m does the line y=m(x+1)
(a) have 2 points of intersection?
(b) have one point of intersection?
(c) have no points of intersection?

I think once I understand the first one I may be able to understand the others, but I have no idea how to do the first one. Help please!!

Answer 1

y = 2x²
and
y = mx + m

therefore
2x² = mx + m

2x² - mx - m = 0
Quadratic formula:

x = [m +/- SQRT(m^2 +8m)] / 4

Solving the above system gives us the intersection points of the parabola and the line.

We control the value of m.

The key to how many solutions there are is all in the square root.

if m^2 + 8m is negative, then the square root does not exist and there are no points of intersection.

if m^2 + 8m is zero, then there is only one point of intersection. It is located at x = m/4

if m^2 + 8m is positive and greater than zero (in math, this is called "strictly positive"), then the square root can have two values (plus or minus) and there are two intersection points.