## Sketching graphs in R3

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### Sketching graphs in R3

Sketch the graph $f(x, y) = y^2 - x^2$.

Let z = 0:

$y^2 - x^2 = k$

Now when drawing the graph on the y and x axis, I understand that there are hyperbolas which intersect y-axis (as letting $x = 0$ you get $y^2 = k$).

What I don't get is in my book their hyperbolas drawn, (on the same graph), which intersect the x-axis. This shouldn't be possible since letting $y = 0$ you get $-x^2 = k$, which can't be solved.
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SyNtHeSiS
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### Re: Sketching graphs in R3

Your question is confusing. Where is z? Also k=0 is possible - that's how you get the lines crossing the origin.
mathman
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### Re: Sketching graphs in R3

What I meant to say is when you let $z = k$ (to get the view from above in R3) you get:

$y^2 - x^2 = k$

$\frac{y^2 - x^2}{k} = 1$

This would be graph of the hyperbolas intersecting the y-axis, but in the book they are hyperbolas which also intersect the x-axis. Don't understand why there are hyperbolas that intersect the x-axis (as letting $y = 0$, you get $-x^2 = k$which can't be solved).
SyNtHeSiS
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### Re: Sketching graphs in R3

The hyperbolas intersecting the x-axis represent negative values of k.

When 0 < k they intersect the y-axis, when k < 0 they intersect the x-axis, and when k = 0, you get degenerate cases, the lines y = ±x.

This is the family of curves satisfying:

$\frac{dy}{dx}=\frac{x}{y}$
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### Re: Sketching graphs in R3

For k < 0, how can you draw $\frac{x^2 - y^2}{k} = -1$, when the right-hand-side has a negative 1? (The standard equation of a hyperbola has a positive 1 on the right-hand-side).
SyNtHeSiS
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### Re: Sketching graphs in R3

The hyperbolas of the form:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

have vertices at (-a,0) and (a,0), while the hyperbolas of the form:

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

have vertices at (0,-a) and (0,a).

In our case, we have $a^2=b^2=k$, so we have:

$y^2-x^2=k$ or $x^2-y^2=-k$

When k < 0, we may choose the form:

$x^2-y^2=-k$

so that now we have a positive value on the right and vertices on the x-axis.

When 0 < k, we choose the form:

$y^2-x^2=k$

so that we have a positive value on the right and vertices on the y-axis.

In both cases, the asymptotes are the lines $y=\pm x$.

When k = 0, we have:

$y^2-x^2=0$

$y=\pm x$
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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