I have totally forgotten how to draw a hyperbola as $3x^2-y^2-4x+1=0$ , I tried to give a look in my old math books but I had not found anything similar to the form of the hyperbola I proposed. Can someone help me to understand in a simple and fine way how can I draw it? (I feel quite embarrassed...)
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It's "hyperbola." "Hyperbole" is something different. :-) – Josephine Moeller Feb 15 '14 at 14:52
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got it, thanks :-) – Dipok Feb 15 '14 at 14:53
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You should start by completing the square to get all of the $x$-terms together, aiming for a form like $\left(\frac{x-x_0}{a}\right)^2$ for those terms. – MPW Feb 15 '14 at 14:55
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If I've done the work correctly, I think you can write this equation as $$\left(\frac{x-\frac23}{3}\right)^2 - \left(\frac{y}{\frac{\sqrt{3}}{3}}\right)^2=1.$$
This shows you can get the hyperbola by linearly transforming each of the variables in the simpler hyperbola $x^2 -y^2=1$. Do you know the shape of that simpler hyperbola? If so, the you can stretch and shift it to get your hyperbola.
MPW
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@MatthewConroy: Yep, just the sort of typo I was afraid of. Corrected. Of course, should have been "-" instead of "+". Thanks for the keen eye. +1 for you. (Oddly enough, I seem to have been upvoted despite that.) – MPW Feb 15 '14 at 22:58
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Drawing the asymptotes and the vertices will get you closest to a graph. First, complete the square so that you have the hyperbola in a more standard form:
$$ (3x - 2)^2 - 3y^2 = 1 $$
The asymptotes will come from replacing the constant on the right by $0$: $(3x - 2)^2 - 3y^2 = 0$, or $y = \pm(3x - 2)/\sqrt{3}$. Draw those lines first.
Then draw the vertices. These will be to the right and left of the asymptotes because the squared $y$ term is negative. They'll also have the same $y$ coordinate as where the asymptotes cross ($y = 0$). That leaves $3x - 2 = \pm 1$ or $x = \{1/3,1\}$.
EDIT: A couple more points that will help nail down the graph are where $x = 0$. That will give you $(0,\pm 1)$. Then you can plug the $y$ values back into the formula to get $9x^2 - 12x = 0$ or $3x(3x - 4) = 0$. Because $3x - 4$ can now also be $0$, this gives you two additional points $(4/3,\pm 1)$.
Josephine Moeller
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$$3x^2-y^2-4x+1=0\iff3x^2-4x+1=y^2\iff y(x)=\pm\sqrt{3x^2-4x+1}.$$ Can you plot it now ?
Lucian
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