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When dealing with quadratic equations, completing the square is a useful technique to rewrite the equation in a form that is easier to interpret or solve. Let's break down how it's done using the original hyperbola equation:
\(2y^{2} - x^{2} - 12y - 6 = 0\).
Firstly, we moved the constant term \(-6\) to the right side of the equation, leading to:

• \(2y^{2} - 12y - x^{2} = 6\)

Next, we focus on the \(y\)-terms, factoring out the coefficient of \(y^{2}\) which is 2:

• \(2(y^{2} - 6y) - x^{2} = 6\)

To complete the square for \(y\), we calculate \((-\frac{6}{2})^2 = 9\). We add and subtract this value inside the parentheses:

• \(2((y - 3)^{2} - 9) - x^{2} = 6\)

Simplifying gives us:

• \((y - 3)^2 - \frac{x^2}{2} = 12\)

Completing the square transforms a quadratic into a perfect square trinomial, critical when converting to standard forms.

The standard form of a hyperbola helps in easily identifying its properties, such as the center, axes orientation, and direction of opening. Once the equation is simplified through completing the square, it’s expressed as:
\[ \frac{x^2}{24} - \frac{(y - 3)^2}{12} = 1 \]Here is how we transition to this form from the completed square step.
After completing the square, we added 18 on both sides to isolate the squared terms, leading to:

• \(2(y - 3)^2 - x^2 = 24\)

Dividing every term by 24 helps us normalize the equation to the standard form:

• \(\frac{x^2}{24} - \frac{(y - 3)^2}{12} = 1\)

In this equation, when \(x\)-term is positive, the hyperbola opens left and right. The center is \((0, 3)\). The transformation into standard form allows quick identification of opening directions and center.

Understanding the vertices and foci of a hyperbola is crucial in sketching its graph, as these points define its shape and orientation.
Vertices are determined from the standard form. For the equation:
\[ \frac{x^2}{24} - \frac{(y - 3)^2}{12} = 1 \]The vertices lie along the transverse axis and are given by \((0, 3 \pm \sqrt{12})\) start here calculation explanation:

• \(\sqrt{12} = 2\sqrt{3} \), hence vertices are \((0, 3 \pm 2\sqrt{3})\)

To find the foci, apply the formula for hyperbolas \(c^2 = a^2 + b^2\), where \(a = \sqrt{24} = 2\sqrt{6}\) and \(b = \sqrt{12} = 2\sqrt{3}\).

• \(c^2 = 24 + 12 = 36\)
• \(c = 6\), so foci are located at \((\pm 6, 3)\)

Vertices and foci are key markers in graphing the hyperbola, providing bounds and focus to accurately depict its path and orientation.