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Completing the square is a method used to transform a quadratic equation into a form that reveals important properties of the graph. This method is essential when dealing with conic sections like hyperbolas. To complete the square for a term such as \(x^2 + bx\), we want to rewrite it as \((x-h)^2\), where \(h\) is chosen such that the equation represents a perfect square trinomial.

For our equation, \(-100(x^2 - 10x)\), we begin by taking the coefficient of \(x\), which is \(-10\), and divide it by 2, resulting in \(-5\). Squaring \(-5\) gives us 25. We add and subtract this number inside the parentheses:

• \[ -100(x^2 - 10x + 25 - 25) \]
• \[ -100((x - 5)^2 - 25) \]

Similarly, for the \(y\) terms, take the coefficient of \(y\), which is \(-10\), divide by 2 to get \(-5\), and square it to get 25. Adding and subtracting it gives:

• \((y^2 - 10y + 25 - 25)\)
• \[((y - 5)^2 - 25)\]

Completing the square helps us in converting the equation into a recognizable form which is crucial for graphing hyperbolas correctly.

The equation of a hyperbola represents the algebraic relationship that describes its shape and orientation in the coordinate plane. In standard form, a hyperbola looks like \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) for vertical hyperbolas, or \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for horizontal hyperbolas.

In our example, after completing the square and rearranging, we transform the initial equation into the form \[\frac{(y-5)^2}{100} - \frac{(x-5)^2}{1} = 1\]

• Here, the minus sign between terms indicates a hyperbola.
• The positive term \((y-5)^2\) suggests the hyperbola is vertical.
• The expressions \((y-k)^2\) and \((x-h)^2\) show the translations along the axes.
• \((h, k) = (5, 5)\) sets the center of the hyperbola.

Understanding the structure of the equation helps in recognizing and predicting the behavior and shape of the hyperbola on a graph.

Vertices and foci are critical elements in the anatomy of a hyperbola. They provide clues about the hyperbola's spread and exact position in the Cartesian plane.

For a vertical hyperbola like \(\frac{(y-5)^2}{100} - \frac{(x-5)^2}{1} = 1\), the formula for vertices is \((h, k \pm a)\), while the formula for the foci is \((h, k \pm c)\). The distance \(a\) represents half the length between the vertices along the axis of symmetry, and \(c\) is the distance from the center to each focus:

• The vertices are found by identifying \(a\), where \(a = \sqrt{100} = 10\). So, the vertices are at \((5, 5 \pm 10) = (5, -5)\) and \((5, 15)\).
• The foci require computing \(c = \sqrt{a^2 + b^2} = \sqrt{101}\). Thus, the foci are \((5, 5 \pm \sqrt{101})\).

Calculating these points is essential for sketching an accurate graph of a hyperbola. They help set bounds for the shape and provide focal points that characterize its ellipse nature.

The standard form of a hyperbola is vital for graphing and understanding its geometrical properties and characteristics. Transforming an equation into this form involves completing the square and isolating terms.
For a vertical hyperbola such as our equation, we aim to express it as \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\). Once the equation is in this standard form, several attributes of the hyperbola are easily identifiable:

• Signs indicate whether the hyperbola opens vertically or horizontally.
• \((h, k)\) represents the center of the hyperbola.
• \(a^2\) and \(b^2\) link to distances within the graph, \(a\) pertains to vertices and \(b\) relates to the conjugate axis.

This structure enables streamlined calculation of dimensions and helps locate vertices, foci, and axes. By dividing through by constants, we obtain a clear, concise equation, essential for interpreting and sketching the hyperbola's graph.