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

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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.
 

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  • #2
SammyS
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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:
[itex]\displaystyle
4x^2-y^2=c-1\,,[/itex]​
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:
[itex]\displaystyle
(2x-y)(2x+y)=c-1\ .[/itex]​
What must the value of c be to make the solution to this be two lines?
 
  • #3
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If I set z equal to 1 the solution is y=2x and y=-2x. Thank you!
 

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