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December 9th, 2011, 09:15 AM   #1
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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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 Hyperbola.jpg (25.3 KB, 403 views)

 December 9th, 2011, 01:47 PM #2 Global Moderator   Joined: May 2007 Posts: 6,216 Thanks: 493 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.
 December 10th, 2011, 01:10 AM #3 Newbie   Joined: Apr 2010 Posts: 12 Thanks: 0 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).
 December 10th, 2011, 01:29 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs 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: $\frac{dy}{dx}=\frac{x}{y}$
 December 16th, 2011, 02:47 AM #5 Newbie   Joined: Apr 2010 Posts: 12 Thanks: 0 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).
 December 16th, 2011, 06:03 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs 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: $y^2-x^2=0$ $y=\pm x$

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