How can you easily graph the trace of z=y²-x²?

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In equation z=y^2-x^2 when graphing the trace for y^2-x^2=k I see we have y=+-x for k=0 else y=+-sqrt(k+x^2) is there a simple way to graph this
 
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We'll to answer my own question I find it convenient to first take k=0 then we have y=+-x then taking k=1 and y=0 we have points (+-1,0) for k<0 then taking x=0 we have point (0,+-1) for k>0 now here we see that as y to +- infinity y will tend to the line y=+-x using lhopatols rule any thoughts here
 
nameVoid said:
In equation z=y^2-x^2 when graphing the trace for y^2-x^2=k I see we have y=+-x for k=0 else y=+-sqrt(k+x^2) is there a simple way to graph this
In addition to the horizontal cross sections you are plotting, it would be useful to plot the traces in the three coordinate planes. For example, in the y-z plane (when x = 0) you get z = y2, a parabola. In the x-z plane, you also get a parabola.
 
Yes I am just interested in that particular trace is there's better way to graph it
 
For y = ##\sqrt{x^2 + k}##, just plot some points. The graph of y = ##-\sqrt{x^2 + k}## is the reflection across the x-axis of the first one.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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