Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
    
en.wikipedia.org/wiki/Hyperboloid – Andrew Dec 7 '12 at 0:42
up vote 2 down vote accepted

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

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


share|cite|improve this answer

I think one of the best way is to intersect $f(x,y,z)=0$ with some planes. I am doing that by using Maple:

While $f=0$ intersects with $z=1,2,3,\cdots,15$, the following shapes are created:

 [> with(plots);
    with(student);
    f := (1/16)*x^2-(1/9)*y^2-z^2 = 1;
    for i to 15 do a[i] := subs(z = i, f) end do;
    implicitplot([seq(a[i], i = 1 .. 15)], x = -45 .. 45, y = -45 .. 45);

enter image description here

While $f=0$ intersects with $x=1,2,3,\cdots,15$, the following shapes are created:

 [> with(plots);
    with(student);
    f := (1/16)*x^2-(1/9)*y^2-z^2 = 1;
    for i to 15 do a[i] := subs(x = i, f) end do;
    implicitplot([seq(a[i], i = 1 .. 15)], z = -45 .. 45, y = -45 .. 45);

enter image description here

And finally, we have the following curves while intersecting $y=1,2,3,\cdots,15$:

enter image description here

Considering all cases in a $xyz$ system of coordinates we get:

enter image description here

share|cite|improve this answer

It's a hyperbola of two sheets. You can determine this in two ways:

1) Examine the signs and powers of the variables (hyperbola of two sheets has two negative signs on two out of the three variables of degree two).

2) Examine the traces of the equation by setting one of the variables to a constant and then identify the two dimensional equation.

share|cite|improve this answer
    
Perhaps giving an example on how to do steps 1 and 2, or even a partial example, would help the OP more. – Colbi Jun 22 at 18:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.