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Completing the square is a technique used to simplify quadratic expressions and make them more manageable. This method transforms a quadratic expression into a perfect square trinomial, which reveals important features like vertices or centers, making it easier to identify the type of conic section represented by the equation.

To complete the square for a binomial like \(x^2 - 10x\), you take half of the coefficient of \(x\), square it, and then add and subtract this square from the expression. In this example, \((-\frac{10}{2})^2\) equals 25. Thus, \(x^2 - 10x\) becomes \((x-5)^2 - 25\).

For the \(y\) term, we have \(-y^2 + 10y\). Applying the same method gives \((\frac{10}{2})^2 = 25\). This means the expression is rewritten as \(-(y-5)^2 + 25\).

Completing the square allows us to convert a given equation into a recognizable form, revealing its geometric identity, and making further analysis of conic sections more straightforward.

A hyperbola is a specific type of conic section characterized by its two separate curves, called branches. Hyperbolas form when a double cone is sliced perpendicular to its vertical axis. The standard equation for a hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for horizontally oriented hyperbolas.

In our exercise, once we completed the square, the equation \((x-5)^2 - (y-5)^2 = 1\) appeared. It is in the standard form of a hyperbola, confirming that the conic section is indeed a hyperbola with the center at \((5, 5)\).

Hyperbolas have several distinct features:

• **Two branches** that open in opposite directions.
• They have asymptotes, which are lines the branches approach but never touch.
• A center, which is the midpoint between the vertices.
• Hyperbolas also differ from ellipses with a subtraction sign in their equation.

Vertices are key points on conic sections. For a hyperbola, the vertices are the points closest to the center on each branch. To find them, we look at the \(a^2\) value from the standard form of the hyperbola's equation.

In the equation \((x-5)^2 - (y-5)^2 = 1\), we identify \(a^2 = 1\) and \(a = 1\). This value of \(a\) tells us the distance from the center to each vertex in the x-direction because the hyperbola is horizontal.

Thus, starting from the center \((5, 5)\), we move 1 unit left and 1 unit right along the x-axis, resulting in vertices at \((4, 5)\) and \((6, 5)\). These points form the closest points of each branch of the hyperbola to its center.

The foci of a hyperbola are crucial points located inside each branch. They are essential for defining a hyperbola, as the difference in distances from any point on the hyperbola to the foci is constant.

To find the foci for a hyperbola, we use the equation \(c^2 = a^2 + b^2\). Here, \(a\) and \(b\) are derived from the standard hyperbola equation. In our case, both \(a\) and \(b\) are 1. Therefore, \(c^2 = 1 + 1\) implying \(c = \sqrt{2}\).

Starting from the center at \((5, 5)\), we move \(\sqrt{2} \approx 1.41\) units left and right along the x-axis, giving us the foci at \((5-\sqrt{2}, 5)\) and \((5+\sqrt{2}, 5)\). These are pivotal points around which the branches of the hyperbola extend.