# System of Equations

1. Sep 24, 2010

1. The problem statement, all variables and given/known data
Shown in the figure is the graph of x = y^2 and a line of slope m that passes through the point (4, 2). Find the value of m such that the line intersects the graph only at (4, 2) and interpret graphically.

2. Relevant equations
x = y^2
y = mx + b

3. The attempt at a solution
Since this is a system of two equations I went ahead and plugged x into y = mx + b. This gives me the equation y = m(y^2) + b. I then get by^2 - y + b = 0. I can plug this into the quadratic formula (or so I think). This gives me y = -(1 +- sqrt(1 - 4b^2)) / 2. I am not sure where to go from here.

2. Sep 24, 2010

### QuarkCharmer

I think you might be overcomplicating this one.

Graph out the function provided, then determine the slope of a line through the given point that would satisfy the conditions. The slope of your line is what is important to determining whether or not it will touch more than 1 location on the function's curve.

Do you have the "figure" that is mentioned in the instructions? That might make understanding the problem a bit easier.

Last edited: Sep 24, 2010
3. Sep 24, 2010

I do have the figure. It is simply a graph of x = y^2 and another line intersecting x = y^2 at (4, 2). I am given no other information on the line that is intersected x = y^2. I am instructed to find the slope of the line so that it only intersects x = y^2 at (4, 2). I have the answer to this problem, but do not know how to get it. The only hints I got from my instructor dealt with using the quadratic formula to solve for x and then for m.

4. Sep 25, 2010

### HallsofIvy

Yes, you are given "other information"! You are told that the line intersects the parabola only at (4, 2)- there are no other intersections. If you look closely at your graph you should see that the only line that intersects the parabola at (4, 2) and no where else is the tangent line to the parabola there.

You could find the slope of the tangent line by differentiating the function but since this is "PreCalculus", you probably are not supposed to and you don't have to. Here is Fermat's method for finding tangents that predates Calculus:

Saying that y= mx+ b (so x= y/m- b/m) intersects $x= y^2$ means, of course, that $x= y^2= y/m- b/m$ or $y^2- (1/m)y+ b/m= 0$. That's the quadratic equation you want to solve.

And saying that the line intersects the parabola only at that point means that the quadratic equation has only one solution- in fact, you know that that one solution must be y= 2. So you must have $y^2- (1/m)y+ b/m= (y- 2)^2$. Expand the right side and set the coefficients equal to find m and b.

(Here, I am not saying you should use the quadratic formula- but you could: the quadratic equation $ax^2+ bx+ c= 0$ has a double root if and only if its "discriminant", $\sqrt{b^2- 4ac}$, is 0.)

5. Sep 25, 2010

That is a great reply. I've gotten a little further, but am stuck again. I used the Quadratic Formula to solve this and came up with y = (-1/m +- sqrt((1/m^2) - (4b - m))) / 2. I know that the problem has only one solution so sqrt((1/m^2) - (4b - m)) = 0. I am not sure what to do from this point.

Also, I am not understanding how you got y^2- (1/m)y+ b/m= (y- 2)^2. Is there a law or property concerning this? I also do not know what it means to set the coefficients equal. This is my first math course in over 6 years. I have forgotten quite a bit. I tried to look in my textbook for any hints or clues, but there are no examples that deal with this specific kind of problem. I am sure I can use the basics from other examples to deal with this, but I simply do not know how.

Last edited: Sep 25, 2010