# Finding a particular level curve of a function z=f(x,y)

1. Sep 19, 2012

### cdot

1. The problem statement, all variables and given/known data

For the function z=f(x,y)=4x^2-y^2+1 I need to set z to a constant c so that the level curve created by the intersection of f(x,y) with the plane z=c is two intersecting lines. I know that the cross section of the function with x fixed is a parabola opening down and the cross section of the function with y fixed is a parabola opening up (it's a hyperbolic paraboloid).

2. Relevant equations

z=4x^-y^2+1

3. The attempt at a solution

I've graphed the function on wolfram alpha and I'm trying to visualize slicing the surface with horizontal planes but I can't seem to visualize at what value of z you would get two intersecting lines as a level curve.

2. Sep 19, 2012

### SammyS

Staff Emeritus
If z = c, your equation becomes equivalent to:
$\displaystyle 4x^2-y^2=c-1\,,$​
which for most values of c is the equation of a pair of hyperbolas.

The left hand side is the difference of squares so the equation is equivalent to:
$\displaystyle (2x-y)(2x+y)=c-1\ .$​
What must the value of c be to make the solution to this be two lines?

3. Sep 20, 2012

### cdot

If I set z equal to 1 the solution is y=2x and y=-2x. Thank you!