We aim to parameterize the surface defined by the hyperboloid equation:

$5x^2 - 5y^2 - z^2 = 5,$

restricted to the part of the surface in front of the $yz$-plane ($x \geq 0$).

Step 1: Rearrange the equation for simplicity

Divide through by 5 to get: $x^2 - y^2 - \frac{z^2}{5} = 1.$

This is a hyperboloid of one sheet. We'll parameterize $y$ and $z$ using hyperbolic trigonometric functions, but for simplicity, we'll directly use sine and cosine.

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Step 2: Define the parameterization

Introduce parameters $u$ and $v$:

• Let $x = \cosh(u)$, ensuring $x \geq 0$ for the part in front of the $yz$-plane.
• Let $y = \sinh(u)\cos(v)$, representing a circular-like motion around the $x$-axis in the $y$-direction.
• Let $z = \sqrt{5} \sinh(u)\sin(v)$, scaled appropriately to match the equation.

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Step 3: Verify the parameterization

Substitute these expressions into the hyperboloid equation: $x^2 - y^2 - \frac{z^2}{5}.$

Substitute: $x^2 = \cosh^2(u), \quad y^2 = \sinh^2(u)\cos^2(v), \quad \frac{z^2}{5} = \sinh^2(u)\sin^2(v).$

Using the identity $\cosh^2(u) - \sinh^2(u) = 1$, we simplify: $\cosh^2(u) - \sinh^2(u)\left(\cos^2(v) + \sin^2(v)\right) = \cosh^2(u) - \sinh^2(u) = 1.$

Thus, the parameterization satisfies the hyperboloid equation.

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Final Answer

The parametric equations are: $x = \cosh(u), \, y = \sinh(u)\cos(v), \, z = \sqrt{5}\sinh(u)\sin(v),$ where $u \in (-\infty, \infty)$ and $v \in [0, 2\pi]$.

Let me know if you have questions or want detailed verification steps!

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Relative questions:

1. Why is $\cosh(u)$ used for $x$ to ensure $x \geq 0$?
2. What would change if the part of the surface behind the $yz$-plane was required instead?
3. How does this parameterization simplify verifying the hyperboloid equation?
4. Can other trigonometric functions be used in this parameterization?
5. How would the parameterization change for a hyperboloid of two sheets?

Tip: Always ensure parameterizations respect the restrictions of the domain in the original equation.