Problem 39 - Intermediate Algebra

Question

Identify whether equation, when graphed, will be a parabola, circle, ellipse, or hyperbola. Sketch the graph of equation. If a parabola, label the vertex. If a circle, label the center and note the radius. If an ellipse, label the center. If a hyperbola, label the \(x\) - or \(y\) -intercepts. \(\frac{y^{2}}{9}-\frac{x^{2}}{9}=1\)

Step-by-Step Solution

Answer

The equation is a hyperbola with \( y \)-intercepts at (0, 3) and (0, -3).

Step 1: Identify the equation form

The given equation is \( \frac{y^2}{9} - \frac{x^2}{9} = 1 \). This equation is in the standard form of a hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), where \( a^2 = 9 \) and \( b^2 = 9 \). This confirms that the graph is a hyperbola.

Step 2: Calculate intercepts

Since this is a vertical hyperbola, it opens up and down. Find the \( y \)-intercepts by setting \( x = 0 \). The equation becomes \( \frac{y^2}{9} = 1 \), which solves to \( y^2 = 9 \). So, \( y = 3 \) and \( y = -3 \). These are the \( y \)-intercepts.

Step 3: Sketch the graph

Plot the two \( y \)-intercepts at points \( (0, 3) \) and \( (0, -3) \). Then, draw the asymptotes by substituting back to form the rectangle: lines \( y = \pm 3x \). Finally, sketch the hyperbola opening vertically around these intercepts and asymtotes to complete the graph.

Step 4: Label the graph

Label the \( y \)-intercepts on the graph at \( (0, 3) \) and \( (0, -3) \). There are no \( x \)-intercepts for this hyperbola. Ensure the appearance of the hyperbola aligns with its vertical orientation determined from the intercepts.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that looks like two mirrored curves opening in opposite directions. It's formed when a plane cuts through both halves of a double cone. The standard form equation of a hyperbola will either have differences involving squared terms, set equal to 1 or another constant. Hyperbolas can open either horizontally (left and right) or vertically (up and down), depending on their equation. In our case, the hyperbola opens vertically since the term with \( y^2 \) comes first in the equation: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Hyperbolas have two branches, each mirroring the other with respect to the center. This center is essentially the midpoint of where the two branches appoach towards but never meet. Understanding the basic structure and characteristics of a hyperbola forms the foundation for graphing these intriguing shapes.

Standard Form Equation

The standard form of a hyperbola's equation helps to easily determine its orientation and key features. For hyperbolas, the standard form equation is written as follows:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]or\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]These two formats tell us whether the hyperbola opens vertically or horizontally.

• Vertical Hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
• Horizontal Hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

In the exercise, the equation \( \frac{y^2}{9} - \frac{x^2}{9} = 1 \) corresponds to a vertical hyperbola. By comparing with the standard form, we can see that \( a = 3 \) and \( b = 3 \), providing information about the hyperbola's dimensions and orientation. Knowing the standard form is essential for identifying and graphing hyperbolas correctly.

Graph Sketching

Graph sketching of a hyperbola involves a sequence of steps that accurately represent its features. For a vertical hyperbola like the one we're dealing with, the initial step is identifying critical points called intercepts. Begin by finding the \( y \)-intercepts. When \( x = 0 \) in our main equation, then\( \frac{y^2}{9} = 1 \), which means \( y = 3 \) and \( y = -3 \). These points are essential and are where the hyperbola will cross the y-axis.Next, plot these \( y \)-intercepts at (0, 3) and (0, -3). The next step is to add asymptotes, which, in this case, are lines \( y = \pm 3x \). They act like a guideline, indicating the direction the hyperbola will extend infinitely. With intercepts in place and shapes of asymptotes drawn, sketch the hyperbola to open upward and downward around this framework. Drawing correctly ensures that the sketch aligns with the geometric properties determined by the equation.

Intercepts

Intercepts are specific points where a graph crosses the axes. In our hyperbola example, there are no \( x \)-intercepts because it never crosses the \( x \)-axis. To locate the \( y \)-intercepts, substitute \( x = 0 \) into the equation\( \frac{y^2}{9} - 0 = 1 \). Solving \( \frac{y^2}{9} = 1 \) gives us \( y^2 = 9 \), hence \( y = 3 \) and \( y = -3 \).These values indicate the points (0, 3) and (0, -3) on the y-axis where the hyperbola intersects. It's important to remember that these intercepts are major characteristics that help guide the sketching of the hyperbola. Knowing where the graph interacts with the axes provides insight into its placement and dimension. This understanding of intercepts is not only crucial for hyperbolas but also for other conic sections like circles, ellipses, and parabolas.