Problem 30 - Calculus Single Variable

Question

The center of an ellipse or hyperbola is the point of intersection of its axes of symmetry. Each state whether the graph of the given Cartesian equation is an ellipse or hyperbola. Determine its standard form and center. \(9 x^{2}-y^{2}-y=1 / 2\)

Step-by-Step Solution

Answer

The equation represents a hyperbola centered at \((0, -\frac{1}{2})\).

Step 1: Recognize the Format

The given equation is \(9x^{2} - y^{2} - y = \frac{1}{2}\). Let us first rearrange it into a recognizable conic section format by getting all terms on one side of the equation.

Step 2: Move Terms

Subtract \(\frac{1}{2}\) from both sides to express the equation as a standard conic: \[9x^2 - y^2 - y - \frac{1}{2} = 0.\]

Step 3: Rearrange the Equation

Rearrange the equation to highlight the terms: \[9x^2 - (y^2 + y) = \frac{1}{2}.\] This shows that there is a quadratic in \(y\).

Step 4: Complete the Square

To complete the square for the \(y\) terms, take \(-y^2 - y\), factor out the negative, and complete the square: \[-(y^2 + y + \frac{1}{4} - \frac{1}{4}) = -((y + \frac{1}{2})^2 - \frac{1}{4}).\] Simplify to get: \[-(y + \frac{1}{2})^2 + \frac{1}{4}.\]

Step 5: Substitute the Completed Square

Place the completed square into the equation: \[9x^2 - [(y + \frac{1}{2})^2 - \frac{1}{4}] = \frac{1}{2}.\] Simplifying results in: \[9x^2 - (y + \frac{1}{2})^2 + \frac{1}{4} = \frac{1}{2}.\] Then subtract \(\frac{1}{4}\) from \(\frac{1}{2}\) to yield: \[9x^2 - (y + \frac{1}{2})^2 = \frac{1}{4}.\]

Step 6: Standard Form Conversion

Divide every term by \(\frac{1}{4}\) to simplify: \[36x^2 - 4(y + \frac{1}{2})^2 = 1.\] This is a hyperbola in the form \(Ax^2 - By^2 = 1\), indicating \(A > 0\) and \(B < 0\).

Step 7: Identify the Center

From \(-4(y + \frac{1}{2})^2\), the transformation \(y + \frac{1}{2}\) reveals the center shift in \(y\). Thus the center is at \((0, -\frac{1}{2})\).

Key Concepts

Conic Sections

Conic sections are the curves we get by slicing a double-napped cone with a plane. These curves come in different shapes based on the angle and position of the slice.
They include:

• Circles
• Ellipses
• Parabolas
• Hyperbolas

These four types of curves have distinguishing features and equations. Among these, hyperbolas and ellipses are often mistaken for each other since both involve squares in their equations.
The key difference is in the signs within their standard equations. Ellipses have a plus sign between their squared terms while hyperbolas have a minus sign.

When you notice an equation with terms like \( Ax^2 \pm By^2 = C \), that involves squares of \( x \) and \( y \), you are dealing with either an ellipse or a hyperbola.
Recognizing these signs is crucial for categorizing the conic sections accurately.

Standard Form

The standard form of conic sections is a specific arrangement of equations that makes these shapes easier to identify and work with.
For hyperbolas, the standard equation often takes the form:\[ Ax^2 - By^2 = 1 \] or, if the axes are switched: \[ By^2 - Ax^2 = 1 \].
The important characteristic of a hyperbola's standard form is the minus sign between the squared terms. This highlights the existence of two separate "branches" or curves.

To convert a quadratic equation into its standard form, you need to rearrange and simplify its terms. This includes factoring, simplifying, and sometimes completing the square.
In our example, once we completed the square and simplified, the equation appeared as: \[ 36x^2 - 4(y + \frac{1}{2})^2 = 1 \].
This clearly shows it's in the standard form of a hyperbola, making further analysis and graphing more intuitive.

Completing the Square

Completing the square is a technique often used in algebra to transform a quadratic equation into a form that is easier to analyze, specifically when converting to standard forms.
This method helps in expressing a quadratic in terms of perfect squares plus a constant term. It is crucial for identifying shifts and transformations in conic sections.

To complete the square:

• Group the quadratic and linear terms together.
• Add and subtract the square necessary to create a perfect square trinomial.
• Factor the trinomial into its squared form.

In our equation \(-y^2 - y\), completing the square involved the steps:
Starting from \[-(y^2 + y)\], by adding and subtracting \(\frac{1}{4}\), it becomes \[-((y + \frac{1}{2})^2 - \frac{1}{4})\].
Finally, this transformation allowed the completion of the square to replace part of the original equation, crucial for moving towards a recognizable conic section's standard form.
Understanding this process aids not just in solving specific equations, but in providing a clearer geometric interpretation of the graph.