Question

Trace the surface
x^2/4 + y^2/9 - z^2/4=1
Also describe its sections by the planes x = ±2 , algebraically and geometrically.

Answer 1

Answer on Question #78593 – Math – Analytic Geometry

Question

Trace the surface

$$\frac {x ^ {2}}{4} + \frac {y ^ {2}}{9} - \frac {z ^ {2}}{4} = 1$$

Also describe its sections by the planes $x = \pm 2$ , algebraically and geometrically.

Solution

One-sheeted hyperboloid.

To determine the $xy$ -trace, set $z = 0$

$$\frac {x ^ {2}}{4} + \frac {y ^ {2}}{9} = 1 \quad \text{the } xy \text{-trace is the ellipse}$$

To determine the $xz$ -trace, set $y = 0$

$$\frac {x ^ {2}}{4} - \frac {z ^ {2}}{4} = 1 \quad \text{the } xz \text{-trace is the hyperbola}$$

To determine the $yz$ -trace, set $x = 0$

$$\frac {y ^ {2}}{9} - \frac {z ^ {2}}{4} = 1 \quad \text{the } yz \text{-trace is the hyperbola}$$

The section of the hyperboloid by the plane $x = -2$

$$\frac {(- 2) ^ {2}}{4} + \frac {y ^ {2}}{9} - \frac {z ^ {2}}{4} = 1$$ $$\frac {y ^ {2}}{9} - \frac {z ^ {2}}{4} = 0$$ $$\left(\frac {y}{3} - \frac {z}{2}\right) \left(\frac {y}{3} + \frac {z}{2}\right) = 0$$

Two lines: $y = -\frac{3}{2} z$ and $y = \frac{3}{2} z$ .

The section of the hyperboloid by the plane $x = 2$

$$\frac {2 ^ {2}}{4} + \frac {y ^ {2}}{9} - \frac {z ^ {2}}{4} = 1$$ $$\frac {y ^ {2}}{9} - \frac {z ^ {2}}{4} = 0$$ $$\left(\frac {y}{3} - \frac {z}{2}\right) \left(\frac {y}{3} + \frac {z}{2}\right) = 0$$

Two lines: $y = -\frac{3}{2} z$ and $y = \frac{3}{2} z$ .

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