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

[cdot]

Homework Statement

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).

Homework Equations

z=4x^-y^2+1

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.

 

[SammyS]

Homework Statement

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).

Homework Equations

z=4x^2-y^2+1

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.

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?

 

[cdot]

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