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I'm stuck trying to figure out which type of quadric surface this equation is:

$$\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$$

I have narrowed it down to a hyperboloid, but cannot determine if it is of one or two sheets. I'm guessing it's two sheets because it has all negative signs, but I'm not sure. Thanks!

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$$\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$$

You're correct: this is an hyperbola, and it does have two sheets. You might want to explore Hyperpoloids for more information on equations of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1,$$ $$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1,$$ $$\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1.$$



$\text{Graph of }\quad\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$

$\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$



Integer solutions: $(x, y, z): (4, 0, 0), (-4, 0, 0)$.

Solutions in $z: z = \pm \dfrac{1}{12}\sqrt{9x^2 - 16y2 -144}$.

Solutions in $y: y = \pm \dfrac34 \sqrt{x^2 - 16z^2 -16}$


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