Problem 22 - Calculus Early Transcendentals

Question

Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. $$9 x^{2}-4 y^{2}=36$$

Step-by-Step Solution

Answer

Center: (0,0); Vertices: (2,0) and (-2,0); Foci: (±√13,0); Asymptotes: y=±3/2x.

Step 1: Identify Hyperbola Equation Standard Form

The general form of the hyperbola equation is \(Ax^2 - By^2 = C\). In this exercise, the given equation is \(9x^2 - 4y^2 = 36\). Divide each term by 36 to rewrite the equation in standard form \(\frac{x^2}{4} - \frac{y^2}{9} = 1\). Thus, the standard form is \(\frac{x^2}{4} - \frac{y^2}{9} = 1\).

Step 2: Identify Key Components of the Hyperbola

From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the center of the hyperbola is \((0, 0)\) as there are no \(h\) or \(k\) terms. For this hyperbola, \(a^2 = 4\) and \(b^2 = 9\), which implies that \(a = 2\) and \(b = 3\). The transverse axis is along the x-axis.

Step 3: Calculate Vertices

The vertices of a hyperbola with its transverse axis along the x-axis are at \((\pm a, 0)\). With \(a = 2\), the vertices are at \((2, 0)\) and \((-2, 0)\).

Step 4: Calculate Foci

To find the foci for the hyperbola, use the formula \(c^2 = a^2 + b^2\). Here, \(c^2 = 4 + 9 = 13\), thus \(c = \sqrt{13}\). The foci are at \((\pm \sqrt{13}, 0)\).

Step 5: Determine Asymptotes

The asymptotes' equations for a hyperbola with its transverse axis along the x-axis are \(y = \pm \frac{b}{a}x\). Here, \(\frac{b}{a} = \frac{3}{2}\), so the asymptotes are \(y = \pm \frac{3}{2}x\).

Step 6: Sketch the Hyperbola

Using the center \((0,0)\), vertices \((2,0)\) and \((-2,0)\), foci \((\pm \sqrt{13}, 0)\), and asymptotes \(y = \pm \frac{3}{2}x\), you can sketch the hyperbola. It will open horizontally, with the vertices on the x-axis and the foci further along the x-axis from the vertices.

Key Concepts

Vertices

The vertices of a hyperbola are key points where the hyperbola intersects its transverse axis. To find these points, you'll need to look at the equation the hyperbola is derived from, which is in the standard form. For instance, in the equation \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \), the term \( a^2 = 4 \) gives us \( a = 2 \). This indicates the distance from the center to each vertex.
In our case, since the transverse axis is on the x-axis, the vertices are at \( (\pm a, 0) \). Substituting \( a \), we find the vertices are at \( (2, 0) \) and \( (-2, 0) \). These vertices serve as the 'endpoints' of the hyperbola's curve in one direction. They help define the overall shape and orientation of the graph.

Foci

The foci ("focus" in the singular form) are points that lie on the transverse axis of a hyperbola. Their role is crucial as they determine how "stretched" or open the hyperbola is. To locate the foci, use the formula \( c^2 = a^2 + b^2 \). In the given equation \( a^2 = 4 \) and \( b^2 = 9 \). Calculating the sum, \( c^2 = 13 \), therefore, \( c = \sqrt{13} \).
The foci are located at \( (\pm \sqrt{13}, 0) \). They will always lie further out than the vertices on the transverse axis, impacting the steepness of the hyperbola's 'arms'. Each point of the hyperbola's branches tends to keep specific continuous proximity to these foci.

Asymptotes

Asymptotes are invisible lines that become apparent in the behavior of hyperbola's arms, acting as boundary guides for how the arms will open. They form a 'box' around the hyperbola, guiding its infinite stretch. To derive the asymptote equations, use the formula \( y = \pm \frac{b}{a}x \). Here, \( \frac{b}{a} = \frac{3}{2} \), which gives us the lines \( y = \pm \frac{3}{2}x \).
These lines are essential because they provide a framework within which the hyperbola grows and extends. They cross the center of the hyperbola and act as tangents at infinity, helping in sketching the hyperbola accurately.

Standard Form

The standard form of a hyperbola is essential for analyzing and sketching it. It allows easy identification of axis length, vertices, and foci. It is represented as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for hyperbolas that open horizontally, or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \) for vertical ones.
Starting from the given equation \( 9x^2 - 4y^2 = 36 \), you transform it to standard form by dividing all terms by 36, resulting in \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \). This form instantly tells us:

• \( a^2 = 4 \) and \( b^2 = 9 \)
• The transverse axis is horizontal, due to the \( x^2 \) term coming first

Understanding the standard form ensures you're prepared to identify all critical components to construct the graph.

Graph Sketching

Sketching a hyperbola involves organizing all the information about its structure onto a graph. Here's how to do it efficiently:

• First, plot the center at \( (0,0) \).
• Next, locate and mark the vertices at \( (2, 0) \) and \( (-2,0) \).
• Then, mark the foci at \( (\pm \sqrt{13}, 0) \).
• Draw the asymptotes by plotting lines through the center with slope \( \pm \frac{3}{2} \).
• Finally, sketch the hyperbola arcs opening horizontally, approaching the asymptotes as they extend away from the center.

Through these steps, you provide a visual representation of the hyperbola based on mathematical calculations, making it an effective way to understand its properties and structure.